Algebra

356 soru

Soru 321Soru

A startup hired two freelance software developers, Developer A and Developer B, for a combined total of 4545 hours on a single project. Developer A charges $65\$65 per hour, and Developer B charges $80\$80 per hour. If the total amount paid to both developers was $3225\$3{}225, how many hours did Developer A work on the project?

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Cevap: 25

Cevap

Developer A worked on the project for 25 hours.
Let xx be the hours Developer A worked and yy be the hours Developer B worked. From x+y=45x + y = 45, we get y=45xy = 45 - x. Substituting into 65x+80y=322565x + 80y = 3225 yields 65x+80(45x)=322565x + 80(45 - x) = 3225. Expanding gives 65x+360080x=322565x + 3600 - 80x = 3225, which simplifies to 15x=375-15x = -375, so x=25x = 25.

Adım Adım Çözüm

1
Set up a system of linear equations representing the total hours and total cost.
x+y=45x + y = 45 and 65x+80y=322565x + 80y = 3225, where xx is Developer A's hours and yy is Developer B's hours.
Translating the verbal conditions into mathematical equations.
2
Express yy in terms of xx from the hours equation.
y=45xy = 45 - x
Prepares the linear system for substitution.
3
Substitute y=45xy = 45 - x into the total cost equation and solve for xx.
65x+80(45x)=3225    15x=375    x=2565x + 80(45 - x) = 3225 \implies -15x = -375 \implies x = 25
Solves for the requested variable xx directly.

Anahtar Kavram

Solving word problems using 2x2 systems of linear equations via substitution or elimination.
Soru 322Soru

For all real numbers xx and yy such that xy|x| \neq |y| and x2+y20x^2 + y^2 \neq 0, consider the algebraic expression:

P(x,y)=(x3+y3x2yxy2x4y4)(x3+y3x2xy+y2)P(x, y) = \left(\frac{x^3 + y^3 - x^2 y - xy^2}{x^4 - y^4}\right) \cdot \left(\frac{x^3 + y^3}{x^2 - xy + y^2}\right)

Which of the following expressions are equivalent to P(x,y)P(x, y) for all valid values of xx and yy? Indicate all such expressions.

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Cevap: x4y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2}; 12y2x2+y21 - \frac{2y^2}{x^2 + y^2}

Cevap

The expressions equivalent to P(x,y)P(x, y) are x4y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2} and 12y2x2+y21 - \frac{2y^2}{x^2 + y^2}.
The given expression simplifies to x2y2x2+y2\frac{x^2 - y^2}{x^2 + y^2}. The option x4y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2} simplifies directly to x2y2x2+y2\frac{x^2 - y^2}{x^2 + y^2} after factoring the numerator. The option 12y2x2+y21 - \frac{2y^2}{x^2 + y^2} simplifies to x2+y22y2x2+y2=x2y2x2+y2\frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2} when combined over a common denominator.

Adım Adım Çözüm

1
Factor the numerator of the first rational term by grouping terms.
x3+y3x2yxy2=x2(xy)y2(xy)=(x2y2)(xy)=(xy)2(x+y)x^3 + y^3 - x^2 y - xy^2 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x - y)^2 (x + y)
Grouping allows rewriting four polynomial terms into product of linear/quadratic factors.
2
Factor the denominator of the first rational term as a difference of squares.
x4y4=(x2y2)(x2+y2)=(xy)(x+y)(x2+y2)x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)
Decomposing x4y4x^4 - y^4 into simpler factors reveals common factors with the numerator.
3
Simplify the first rational term by canceling common factors (xy)(x+y)(x - y)(x + y).
\frac{(x - y)^2(x + y)}{(x - y)(x + y)(x^2 + y^2)} = \frac{x - y}{x^2 + y^2}
Canceling non-zero common factors simplifies the fraction.
4
Factor the numerator of the second rational term using the sum of cubes formula.
x3+y3=(x+y)(x2xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2)
Applying the sum of cubes identity exposes the irreducible quadratic factor present in the denominator.
5
Simplify the second rational term and multiply the result by the simplified first term.
P(x,y)=(xyx2+y2)(x+y)=(xy)(x+y)x2+y2=x2y2x2+y2P(x, y) = \left(\frac{x - y}{x^2 + y^2}\right) \cdot (x + y) = \frac{(x - y)(x + y)}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}
Multiplying the simplified forms yields the simplest explicit representation of P(x,y)P(x, y).
6
Verify equivalence of the options against x2y2x2+y2\frac{x^2 - y^2}{x^2 + y^2}.
The option x4y4(x2+y2)2=(x2y2)(x2+y2)(x2+y2)2=x2y2x2+y2\frac{x^4 - y^4}{(x^2 + y^2)^2} = \frac{(x^2 - y^2)(x^2 + y^2)}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{x^2 + y^2}, and the option 12y2x2+y2=x2+y22y2x2+y2=x2y2x2+y21 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}. Both match P(x,y)P(x, y).
Transforming algebraic expressions under common denominators or factoring confirms equivalence.

Anahtar Kavram

Simplifying complex algebraic expressions using factoring by grouping, difference of squares, sum of cubes, and common denominator manipulation.
Soru 323Soru

A bakery packages two types of gift baskets containing gourmet croissants and blueberry muffins. Basket X contains 44 croissants and 33 muffins, with a total production cost of $19.00\$19.00. Basket Y contains 22 croissants and 55 muffins, with a total production cost of $16.50\$16.50. Assuming the cost per croissant and the cost per muffin are constant across all baskets, what is the production cost, in dollars, of a single croissant?

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Cevap: 3.25

Cevap

The production cost of a single croissant is 3.25 dollars.
Let cc represent the cost of a croissant and mm represent the cost of a muffin. The given situation translates to the system of equations 4c+3m=19.004c + 3m = 19.00 and 2c+5m=16.502c + 5m = 16.50. Multiplying the second equation by 22 gives 4c+10m=33.004c + 10m = 33.00. Subtracting 4c+3m=19.004c + 3m = 19.00 from 4c+10m=33.004c + 10m = 33.00 results in 7m=14.007m = 14.00, which gives m=2.00m = 2.00. Substituting m=2.00m = 2.00 into 2c+5(2.00)=16.502c + 5(2.00) = 16.50 yields 2c+10=16.502c + 10 = 16.50, so 2c=6.502c = 6.50 and c=3.25c = 3.25. Thus, a single croissant costs $3.25 dollars.

Adım Adım Çözüm

1
Set up a system of linear equations
4c+3m=19.004c + 3m = 19.00 and 2c+5m=16.502c + 5m = 16.50
Translate the contents and costs of Basket X and Basket Y into algebraic equations where cc is the price of a croissant and mm is the price of a muffin.
2
Eliminate variable c
4c+10m=33.004c + 10m = 33.00, then subtracting 4c+3m=19.004c + 3m = 19.00 yields 7m=14.007m = 14.00, so m=2.00m = 2.00
Multiplying the second equation by 2 aligns the coefficients of cc, allowing elimination by subtraction.
3
Solve for variable c
2c+5(2.00)=16.50    2c=6.50    c=3.252c + 5(2.00) = 16.50 \implies 2c = 6.50 \implies c = 3.25
Substitute the value found for mm back into one of the original linear equations to calculate the cost of a croissant.

Anahtar Kavram

Solving Systems of Linear Equations via Elimination

Alternatif Yöntem

Express cc in terms of mm using the second equation: c=8.252.5mc = 8.25 - 2.5m. Substitute this expression into the first equation: 4(8.252.5m)+3m=19.00    3310m+3m=19.00    7m=14.00    m=2.004(8.25 - 2.5m) + 3m = 19.00 \implies 33 - 10m + 3m = 19.00 \implies -7m = -14.00 \implies m = 2.00. Finally, calculate c=8.252.5(2.00)=3.25c = 8.25 - 2.5(2.00) = 3.25.
Tahmini Süre:1m 30s
Soru 324Soru
For all real numbers xx and yy such that xy|x| \neq |y|, which of the following expressions is equivalent to x3+x2y+x2y3xy2y2x2y2?\frac{x^3 + x^2y + x^2 - y^3 - xy^2 - y^2}{x^2 - y^2}?
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Cevap: x+y+1x + y + 1

Cevap

x+y+1x + y + 1
Factoring the numerator by grouping yields (xy)(x+y)(x+y+1)(x - y)(x + y)(x + y + 1). Factoring the denominator gives (xy)(x+y)(x - y)(x + y). Canceling the non-zero common factor (xy)(x+y)(x - y)(x + y) leaves the simplified expression x+y+1x + y + 1.

Adım Adım Çözüm

1
Group terms in the numerator to identify common factor pairs.
N=(x3y3)+(x2yxy2)+(x2y2)N = (x^3 - y^3) + (x^2y - xy^2) + (x^2 - y^2)
Grouping cubic terms, quadratic cross-terms, and difference of squares separately allows factoring out fundamental algebraic patterns.
2
Apply standard algebraic formulas to each grouped term.
N=(xy)(x2+xy+y2)+xy(xy)+(xy)(x+y)N = (x - y)(x^2 + xy + y^2) + xy(x - y) + (x - y)(x + y)
Using difference of cubes x3y3=(xy)(x2+xy+y2)x^3 - y^3 = (x-y)(x^2+xy+y^2) and difference of squares x2y2=(xy)(x+y)x^2 - y^2 = (x-y)(x+y) reveals a common (xy)(x-y) factor across all terms.
3
Factor out (xy)(x - y) from the numerator and simplify the remaining polynomial.
N=(xy)[(x2+xy+y2)+xy+(x+y)]=(xy)[x2+2xy+y2+x+y]N = (x - y)\left[(x^2 + xy + y^2) + xy + (x + y)\right] = (x - y)\left[x^2 + 2xy + y^2 + x + y\right]
Combining like terms inside the bracket simplifies the expression.
4
Recognize the perfect square trinomial inside the expression.
N=(xy)[(x+y)2+(x+y)]=(xy)(x+y)(x+y+1)N = (x - y)\left[(x + y)^2 + (x + y)\right] = (x - y)(x + y)(x + y + 1)
Rewriting x2+2xy+y2x^2 + 2xy + y^2 as (x+y)2(x + y)^2 allows factoring out (x+y)(x + y).
5
Divide the factored numerator by the denominator.
(xy)(x+y)(x+y+1)(xy)(x+y)=x+y+1\frac{(x - y)(x + y)(x + y + 1)}{(x - y)(x + y)} = x + y + 1
Since xy|x| \neq |y|, both (xy)(x - y) and (x+y)(x + y) are non-zero and can be canceled.

Anahtar Kavram

Factoring multivariable polynomials using grouping, difference of cubes, and difference of squares formulas.
Soru 325Soru
Consider the system of linear equations in two variables xx and yy shown below, where kk is a constant:
2x+ky=103x6y=15\begin{aligned} 2x + ky &= 10 \\ 3x - 6y &= 15 \end{aligned}
Which of the following statements must be true? Select all such statements.

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Cevap: If k=4k = -4, the system has infinitely many solutions.; If k=0k = 0, the system has a unique solution (x,y)=(5,0)(x, y) = (5, 0).

Cevap

The correct statements are the ones stating that if k=4k = -4, the system has infinitely many solutions, and if k=0k = 0, the system has a unique solution (x,y)=(5,0)(x, y) = (5, 0).
Analyzing the simplified relation (k+4)y=0(k + 4)y = 0 demonstrates that setting k=4k = -4 makes the equation identity 0y=00y = 0, yielding infinitely many solutions. For any other value of kk, including k=0k = 0, yy must equal 00, which gives x=5x = 5, establishing a unique solution at (5,0)(5, 0).

Adım Adım Çözüm

1
Simplify the second equation to express xx in terms of yy.
3x6y=15    x2y=5    x=2y+53x - 6y = 15 \implies x - 2y = 5 \implies x = 2y + 5.
Expressing xx explicitly allows direct substitution into the first linear equation.
2
Substitute x=2y+5x = 2y + 5 into the first equation 2x+ky=102x + ky = 10.
2(2y+5)+ky=10    4y+10+ky=10    (k+4)y=02(2y + 5) + ky = 10 \implies 4y + 10 + ky = 10 \implies (k + 4)y = 0.
This reduces the 2x2 system to a single linear equation in yy parameterized by kk.
3
Analyze the conditions for yy based on the parameter kk.
If k=4k = -4, the equation becomes 0y=00y = 0, which is true for all real yy (infinitely many solutions). If k4k \neq -4, then y=0y = 0 and x=5x = 5 (a unique solution).
Determines system consistency and solution multiplicity across all values of kk.

Anahtar Kavram

Parametric Analysis of 2x2 Linear Systems
Soru 326Soru

For pairwise distinct real numbers xx, yy, and zz, consider the algebraic expression:

E(x,y,z)=(x2y2)3+(y2z2)3+(z2x2)3(xy)3+(yz)3+(zx)3E(x, y, z) = \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}

If x=5x = 5, y=3y = 3, and z=1z = 1, what is the numerical value of E(5,3,1)E(5, 3, 1)?

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Cevap: 192

Cevap

192
Using the identity that a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc whenever a+b+c=0a + b + c = 0, both the numerator and denominator can be factored directly. Factoring the difference of squares in the numerator yields 3(xy)(x+y)(yz)(y+z)(zx)(z+x)3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x). Dividing this by the factored denominator 3(xy)(yz)(zx)3(x - y)(y - z)(z - x) simplifies the expression to (x+y)(y+z)(z+x)(x + y)(y + z)(z + x). Substituting x=5x = 5, y=3y = 3, and z=1z = 1 yields (8)(4)(6)=192(8)(4)(6) = 192.

Adım Adım Çözüm

1
Use the conditional cubic identity a+b+c=0    a3+b3+c3=3abca + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc on the denominator.
Denominator becomes 3(xy)(yz)(zx)3(x - y)(y - z)(z - x).
The sum of the three terms (xy)+(yz)+(zx)(x - y) + (y - z) + (z - x) equals 0.
2
Apply the same identity to the numerator.
Numerator becomes 3(x2y2)(y2z2)(z2x2)3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2).
The sum of the squared difference terms (x2y2)+(y2z2)+(z2x2)(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) also equals 0.
3
Factor each difference of squares in the numerator.
Numerator becomes 3(xy)(x+y)(yz)(y+z)(zx)(z+x)3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x).
Using the difference of squares identity u2v2=(uv)(u+v)u^2 - v^2 = (u - v)(u + v) on each term.
4
Simplify the fraction by dividing the common factors in the numerator and denominator.
E(x,y,z)=(x+y)(y+z)(z+x)E(x, y, z) = (x + y)(y + z)(z + x).
The factors 33, (xy)(x - y), (yz)(y - z), and (zx)(z - x) cancel out completely.
5
Evaluate the simplified product for x=5x = 5, y=3y = 3, and z=1z = 1.
(5+3)(3+1)(1+5)=8×4×6=192(5 + 3)(3 + 1)(1 + 5) = 8 \times 4 \times 6 = 192.
Direct evaluation after algebraic simplification.

Anahtar Kavram

Simplifying rational expressions involving sum of cubes identity a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc when a+b+c=0a + b + c = 0 and difference of squares factoring.
Soru 327Soru

A quadratic function is defined by f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are real constants with a>0a > 0. If the vertex of the parabola y=f(x)y = f(x) lies in the third quadrant of the xyxy-plane, which of the following statements must be true? Select all that apply.

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Cevap: b>0b > 0; The equation f(x)=0f(x) = 0 has two distinct real solutions.

Cevap

The statements 'b>0b > 0' and 'The equation f(x)=0f(x) = 0 has two distinct real solutions' must be true.
The vertex of the parabola y=ax2+bx+cy = ax^2 + bx + c is (h,k)=(b2a,cb24a)(h, k) = \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right). Because the vertex is in the third quadrant, h<0h < 0 and k<0k < 0. With a>0a > 0, the inequality b2a<0-\frac{b}{2a} < 0 directly requires b>0b > 0. Furthermore, because a>0a > 0 (parabola opens upward) and the minimum value k<0k < 0 lies below the x-axis, the graph must cross the x-axis at two distinct points, establishing that f(x)=0f(x) = 0 has two distinct real solutions.

Adım Adım Çözüm

1
Analyze the coordinates of the vertex (h,k)(h, k) relative to the third quadrant.
The third quadrant requires h<0h < 0 and k<0k < 0.
Points in Quadrant III have negative x-coordinates and negative y-coordinates.
2
Evaluate the sign of bb using the x-coordinate formula h=b2ah = -\frac{b}{2a}.
b>0b > 0.
Since h<0h < 0 and a>0a > 0, b2a<0    b2a>0    b>0-\frac{b}{2a} < 0 \implies \frac{b}{2a} > 0 \implies b > 0.
3
Determine the number of real solutions using the vertex y-coordinate kk and orientation a>0a > 0.
The discriminant b24ac>0b^2 - 4ac > 0, giving two distinct real solutions.
For an upward-opening parabola (a>0a > 0), having a vertex below the x-axis (k<0k < 0) guarantees the curve crosses the x-axis twice.
4
Test counterexamples for cc and evaluate the root sum ba-\frac{b}{a}.
cc is not strictly constrained in sign, and the root sum is negative.
The function f(x)=(x+2)21=x2+4x+3f(x) = (x+2)^2 - 1 = x^2 + 4x + 3 has vertex (2,1)QIII(-2, -1) \in \text{QIII} with c=3>0c = 3 > 0. Also, since a>0a > 0 and b>0b > 0, root sum ba<0-\frac{b}{a} < 0.

Anahtar Kavram

Parabola Vertex and Quadratic Discriminant Analysis
Soru 328Soru
For all real numbers xx and yy such that xyx \neq y, xyx \neq -y, and x2+y20x^2 + y^2 \neq 0, which of the following expressions is equivalent to
x4y4x3x2y+xy2y32xyx+y\frac{x^4 - y^4}{x^3 - x^2 y + x y^2 - y^3} - \frac{2 x y}{x + y}
Cevabı ve açıklamayı göster

Cevap: x2+y2x+y\frac{x^2 + y^2}{x + y}

Cevap

x2+y2x+y\frac{x^2 + y^2}{x + y}
Factoring x4y4x^4 - y^4 as (xy)(x+y)(x2+y2)(x - y)(x + y)(x^2 + y^2) and x3x2y+xy2y3x^3 - x^2 y + x y^2 - y^3 as (xy)(x2+y2)(x - y)(x^2 + y^2) simplifies the first term to x+yx + y. Combining x+yx + y with 2xyx+y-\frac{2xy}{x+y} over the common denominator (x+y)(x + y) gives (x+y)22xyx+y=x2+y2x+y\frac{(x+y)^2 - 2xy}{x+y} = \frac{x^2 + y^2}{x+y}.

Adım Adım Çözüm

1
Factor the numerator and denominator of the first rational expression
Numerator: x4y4=(x2y2)(x2+y2)=(xy)(x+y)(x2+y2)x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2). Denominator: x3x2y+xy2y3=x2(xy)+y2(xy)=(xy)(x2+y2)x^3 - x^2 y + x y^2 - y^3 = x^2(x - y) + y^2(x - y) = (x - y)(x^2 + y^2).
Factoring allows for cancellation of common factors in rational expressions.
2
Simplify the first rational expression by canceling common terms
(xy)(x+y)(x2+y2)(xy)(x2+y2)=x+y\frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + y^2)} = x + y.
Since xyx \neq y and x2+y20x^2 + y^2 \neq 0, the common terms (xy)(x - y) and (x2+y2)(x^2 + y^2) are non-zero and can be divided out.
3
Subtract the second expression using a common denominator
(x+y)2xyx+y=(x+y)2x+y2xyx+y=(x+y)22xyx+y(x + y) - \frac{2xy}{x + y} = \frac{(x + y)^2}{x + y} - \frac{2xy}{x + y} = \frac{(x + y)^2 - 2xy}{x + y}.
Combining terms under the common denominator (x+y)(x + y) enables algebraic reduction.
4
Expand the squared binomial in the numerator and simplify like terms
\frac{x^2 + 2xy + y^2 - 2xy}{x + y} = \frac{x^2 + y^2}{x + y}.
Expanding (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2 allows the +2xy+2xy and 2xy-2xy terms to cancel out.

Anahtar Kavram

Simplifying rational expressions using polynomial factoring (difference of squares and grouping) and common denominators.
Tahmini Süre:2m 0s
Soru 329Soru

A research laboratory operates two types of automated centrifuges, Model X and Model Y. A single operating cycle of Model X processes 4040 biological samples and consumes 1010 kilowatt-hours (kWh) of electricity. A single operating cycle of Model Y processes 2525 biological samples and consumes 1515 kWh of electricity. On a given day, the laboratory processed a total of 775775 biological samples and consumed 325325 kWh of electricity using only these two models. What is the total number of operating cycles completed by Model X and Model Y combined on that day?

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Cevap: 25

Cevap

The total number of combined operating cycles completed by Model X and Model Y is 25.
The correct answer is 25. Setting up the linear system 40x+25y=77540x + 25y = 775 and 10x+15y=32510x + 15y = 325 and solving via elimination gives x=10x = 10 cycles for Model X and y=15y = 15 cycles for Model Y. Adding these together gives 10+15=2510 + 15 = 25 total cycles.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations based on total samples and total electricity consumption.
Let xx be the number of cycles for Model X and yy be the number of cycles for Model Y.
System of equations:
1) 40x+25y=77540x + 25y = 775 (Sample constraint)
2) 10x+15y=32510x + 15y = 325 (Energy constraint)
Modeling the word problem as two linear equations in two variables allows for algebraic elimination.
2
Simplify both equations by dividing by their greatest common factors.
Divide Equation 1 by 5: 8x+5y=1558x + 5y = 155
Divide Equation 2 by 5: 2x+3y=652x + 3y = 65
Simplifying coefficients reduces computational complexity and minimizes arithmetic errors.
3
Eliminate variable xx by multiplying the simplified second equation by 4 and subtracting the simplified first equation.
4×(2x+3y=65)8x+12y=2604 \times (2x + 3y = 65) \Rightarrow 8x + 12y = 260
Subtract (8x+5y=155)(8x + 5y = 155):
(8x8x)+(12y5y)=260155(8x - 8x) + (12y - 5y) = 260 - 155
7y=105y=157y = 105 \Rightarrow y = 15
Equalizing the coefficients of xx enables solving directly for yy.
4
Substitute y=15y = 15 back into 2x+3y=652x + 3y = 65 to find xx.
2x+3(15)=652x+45=652x=20x=102x + 3(15) = 65 \Rightarrow 2x + 45 = 65 \Rightarrow 2x = 20 \Rightarrow x = 10
Evaluating xx completes the solution for individual cycle counts.
5
Calculate the requested combined total x+yx + y.
x+y=10+15=25x + y = 10 + 15 = 25
The question specifically requests the sum of operating cycles of both models combined.

Anahtar Kavram

Systems of Linear Equations in Two Variables
Tahmini Süre:1m 45s
Soru 330Soru

For all non-zero real numbers xx and yy such that xy|x| \neq |y| and x2+2xy+2y20x^2 + 2xy + 2y^2 \neq 0, consider the algebraic expression:

Q(x,y)=x4+4y4x2+2xy+2y2+2x3y2xy3x2y2Q(x, y) = \frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} + \frac{2x^3 y - 2xy^3}{x^2 - y^2}

Which of the following expressions are equivalent to Q(x,y)Q(x, y)? Indicate all such expressions.

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Cevabı ve açıklamayı göster

Cevap: (xy)2+2xy+y2(x - y)^2 + 2xy + y^2; \frac{x^3 + 2xy^2}{x}

Cevap

The expressions equivalent to Q(x,y)Q(x, y) are (xy)2+2xy+y2(x - y)^2 + 2xy + y^2 and x3+2xy2x\frac{x^3 + 2xy^2}{x}.
First, simplify Q(x,y)Q(x, y) by factoring each term. Using Sophie Germain's identity on the numerator of the first term gives x4+4y4=(x2+2y2)2(2xy)2=(x2+2xy+2y2)(x22xy+2y2)x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2). Dividing this by (x2+2xy+2y2)(x^2 + 2xy + 2y^2) results in x22xy+2y2x^2 - 2xy + 2y^2. For the second term, factoring out 2xy2xy gives 2xy(x2y2)x2y2=2xy\frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy. Adding the two simplified terms yields Q(x,y)=(x22xy+2y2)+2xy=x2+2y2Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2.

Evaluating the options for equivalence:
- The expression (xy)2+2xy+y2(x - y)^2 + 2xy + y^2 expands to x22xy+y2+2xy+y2=x2+2y2x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2.
- The expression x3+2xy2x\frac{x^3 + 2xy^2}{x} factors as x(x2+2y2)x=x2+2y2\frac{x(x^2 + 2y^2)}{x} = x^2 + 2y^2.
Therefore, both of these expressions are equivalent to Q(x,y)Q(x, y).

Adım Adım Çözüm

1
Simplify the first algebraic fraction using Sophie Germain's identity.
x4+4y4x2+2xy+2y2=(x2+2y2)2(2xy)2x2+2xy+2y2=(x2+2xy+2y2)(x22xy+2y2)x2+2xy+2y2=x22xy+2y2\frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2y^2)^2 - (2xy)^2}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)}{x^2 + 2xy + 2y^2} = x^2 - 2xy + 2y^2
Completing the square on x4+4y4x^4 + 4y^4 allows it to be factored into the product of two quadratic expressions.
2
Simplify the second algebraic fraction by factoring out common factors.
2x3y2xy3x2y2=2xy(x2y2)x2y2=2xy\frac{2x^3 y - 2xy^3}{x^2 - y^2} = \frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy
The term (x2y2)(x^2 - y^2) cancels out since x±yx \neq \pm y.
3
Combine the simplified terms to find the closed-form expression for Q(x,y)Q(x, y).
Q(x,y)=(x22xy+2y2)+2xy=x2+2y2Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2
The 2xy-2xy and +2xy+2xy terms sum to zero.
4
Test each option for equivalence to x2+2y2x^2 + 2y^2.
The option (xy)2+2xy+y2(x - y)^2 + 2xy + y^2 expands to x22xy+y2+2xy+y2=x2+2y2x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2. The option x3+2xy2x\frac{x^3 + 2xy^2}{x} simplifies directly to x2+2y2x^2 + 2y^2.
Both expressions reduce identically to x2+2y2x^2 + 2y^2 for all non-zero xx and yy.

Anahtar Kavram

Simplifying complex algebraic expressions using polynomial factoring identities (Sophie Germain identity and difference of squares) and algebraic reduction.
Tahmini Süre:2m 0s
Soru 331Soru

A community theater sold a total of 250250 tickets for a performance, consisting of adult tickets for $15\$15 each and student tickets for $10\$10 each. The total revenue collected from ticket sales was $3,100\$3,100. Let aa represent the number of adult tickets sold and ss represent the number of student tickets sold. Which of the following statements must be true? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The number of student tickets sold was 130130.; The total revenue generated from adult ticket sales was $1,800\$1,800.

Cevap

The correct statements are that the number of student tickets sold was 130, and the total revenue generated from adult ticket sales was $1,800.
The system of equations a+s=250a + s = 250 and 15a+10s=310015a + 10s = 3100 uniquely yields a=120a = 120 adult tickets and s=130s = 130 student tickets. Therefore, the statement that 130 student tickets were sold is true. Furthermore, multiplying 120 adult tickets by their price of 15perticketgivesatotaladultticketrevenueof15 per ticket gives a total adult ticket revenue of 1,800, making that statement true as well.

Adım Adım Çözüm

1
Set up a system of two linear equations in terms of aa and ss.
Equation 1 (total tickets): a+s=250a + s = 250; Equation 2 (total revenue): 15a+10s=310015a + 10s = 3100.
The sum of the ticket quantities equals total tickets, and the sum of monetary contributions equals total revenue.
2
Express ss in terms of aa using Equation 1 and substitute into Equation 2.
s=250a    15a+10(250a)=3100    15a+250010a=3100s = 250 - a \implies 15a + 10(250 - a) = 3100 \implies 15a + 2500 - 10a = 3100.
Using substitution eliminates variable ss to solve for aa.
3
Solve for aa and then find ss.
5a=600    a=1205a = 600 \implies a = 120; s=250120=130s = 250 - 120 = 130.
Dividing 600600 by 55 gives a=120a = 120, and subtracting 120120 from 250250 gives s=130s = 130.
4
Evaluate the statements against the values a=120a = 120 and s=130s = 130.
s=130s = 130 is true. Revenue from adult tickets is 120×15=$1,800120 \times 15 = \$1,800, which is also true.
Comparing calculated values directly confirms which options state accurate quantitative properties.

Anahtar Kavram

Solving 2x2 Systems of Linear Equations using Substitution or Elimination in Word Problems
Soru 332Soru
Consider the system of linear equations in two variables xx and yy shown below:
3x+2y=16x4y=4\begin{aligned} 3x + 2y &= 16 \\ x - 4y &= -4 \end{aligned}
If (x,y)(x, y) is the unique solution to the system, what is the value of yx\frac{y}{x}?
Cevabı ve açıklamayı göster

Cevap: 12\frac{1}{2}

Cevap

The value of yx\frac{y}{x} is 12\frac{1}{2}.
Solving the system by substitution gives x=4x = 4 and y=2y = 2. Dividing yy by xx gives 24=12\frac{2}{4} = \frac{1}{2}, which is the target ratio.

Adım Adım Çözüm

1
Isolate xx in the second equation.
x=4y4x = 4y - 4
Expressing xx in terms of yy allows for substitution into the first equation.
2
Substitute x=4y4x = 4y - 4 into the first equation.
3(4y4)+2y=16    12y12+2y=16    14y=28    y=23(4y - 4) + 2y = 16 \implies 12y - 12 + 2y = 16 \implies 14y = 28 \implies y = 2
Solving the single-variable equation determines the value of yy.
3
Substitute y=2y = 2 back into the isolated expression for xx.
x=4(2)4=4x = 4(2) - 4 = 4
Determines the corresponding value of xx.
4
Compute the required ratio yx\frac{y}{x}.
yx=24=12\frac{y}{x} = \frac{2}{4} = \frac{1}{2}
Evaluates the final expression requested by the prompt.

Anahtar Kavram

Solving systems of linear equations using substitution or elimination to evaluate a combined expression.
Soru 333Soru

A commercial print shop operates two types of high-speed printers, Printer M and Printer N. Operating simultaneously for 55 hours, 22 units of Printer M and 33 units of Printer N print a total of 5,5005,500 pages. Operating simultaneously for 44 hours, 55 units of Printer M and 22 units of Printer N print a total of 6,6006,600 pages. What is the hourly page output of a single Printer M?

Cevabı ve açıklamayı göster

Cevap: 250

Cevap

The hourly page output of a single Printer M is 250 pages per hour.
Dividing each total page output by the corresponding number of hours produces the simplified linear system: 2m+3n=11002m + 3n = 1100 and 5m+2n=16505m + 2n = 1650, where mm and nn are the hourly rates of Printer M and Printer N. Multiplying the first equation by 22 gives 4m+6n=22004m + 6n = 2200, and multiplying the second equation by 33 gives 15m+6n=495015m + 6n = 4950. Subtracting the two equations eliminates nn, giving 11m=275011m = 2750, which simplifies to m=250m = 250 pages per hour.

Adım Adım Çözüm

1
Define variables and write initial algebraic equations based on time and rate.
Let mm represent the hourly page output of Printer M and nn represent the hourly page output of Printer N. The total outputs give 5(2m+3n)=55005(2m + 3n) = 5500 and 4(5m+2n)=66004(5m + 2n) = 6600.
Total page output equals total operating time multiplied by the combined hourly output rate.
2
Simplify the system by dividing each equation by its respective number of hours.
First equation: 2m+3n=11002m + 3n = 1100. Second equation: 5m+2n=16505m + 2n = 1650.
Simplifying yields a standard system of linear equations representing the combined hourly rate.
3
Eliminate variable nn to solve for mm.
Multiply 2m+3n=11002m + 3n = 1100 by 22 to get 4m+6n=22004m + 6n = 2200. Multiply 5m+2n=16505m + 2n = 1650 by 33 to get 15m+6n=495015m + 6n = 4950. Subtracting the first modified equation from the second yields 11m=275011m = 2750, so m=250m = 250.
Equating the coefficients of nn allows direct elimination of nn when subtracting the equations.

Anahtar Kavram

Solving a 2x2 system of linear equations using the method of elimination
Tahmini Süre:1m 30s
Soru 334Soru
For all real numbers x5x \neq -5, the algebraic expression
x3+125x25x+25x225x+5\frac{x^3 + 125}{x^2 - 5x + 25} - \frac{x^2 - 25}{x + 5}
simplifies to a single constant value. What is the value of this constant?
Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The simplified value of the expression for all valid real numbers x is 10.
Factoring the numerators reveals that the first term reduces to x+5x + 5 and the second term reduces to x5x - 5. Subtracting (x5)(x - 5) from (x+5)(x + 5) yields (x+5)(x5)=10(x + 5) - (x - 5) = 10, which is constant for all valid values of xx.

Adım Adım Çözüm

1
Factor the numerator of the first rational term using the sum of cubes identity.
x3+125=(x+5)(x25x+25)x^3 + 125 = (x + 5)(x^2 - 5x + 25)
The sum of cubes formula a3+b3=(a+b)(a2ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2) applies with a=xa = x and b=5b = 5.
2
Cancel the non-zero factor (x25x+25)(x^2 - 5x + 25) from the first fraction.
\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5
The quadratic factor x25x+25x^2 - 5x + 25 has a negative discriminant ((5)24(1)(25)=75<0(-5)^2 - 4(1)(25) = -75 < 0), so it is never zero for any real number xx.
3
Factor the numerator of the second rational term using the difference of squares identity.
x225=(x5)(x+5)x^2 - 25 = (x - 5)(x + 5)
The difference of squares formula a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b) applies with a=xa = x and b=5b = 5.
4
Cancel the common factor (x+5)(x + 5) from the second fraction.
\frac{(x - 5)(x + 5)}{x + 5} = x - 5
Given x5x \neq -5, the factor x+5x + 5 is non-zero and can be canceled.
5
Subtract the two simplified terms.
(x+5)(x5)=x+5x+5=10(x + 5) - (x - 5) = x + 5 - x + 5 = 10
Distribute the negative sign to both terms in (x5)(x - 5) and combine like terms.

Anahtar Kavram

Factoring sum of cubes and difference of squares to simplify rational expressions
Soru 335Soru

For all real numbers x2x \neq 2, which of the following expressions is equivalent to x416x32x2+4x8\frac{x^4 - 16}{x^3 - 2x^2 + 4x - 8}?

Cevabı ve açıklamayı göster

Cevap: x+2x + 2

Cevap

x+2x + 2
Factoring the numerator x416x^4 - 16 as a difference of squares yields (x2)(x+2)(x2+4)(x - 2)(x + 2)(x^2 + 4). Factoring the denominator x32x2+4x8x^3 - 2x^2 + 4x - 8 by grouping yields (x2)(x2+4)(x - 2)(x^2 + 4). Dividing the numerator by the denominator allows the shared factor (x2)(x2+4)(x - 2)(x^2 + 4) to cancel, leaving x+2x + 2.

Adım Adım Çözüm

1
Factor the numerator x416x^4 - 16
x416=(x24)(x2+4)=(x2)(x+2)(x2+4)x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)
Apply the difference of squares factorization identity twice.
2
Factor the denominator x32x2+4x8x^3 - 2x^2 + 4x - 8 by grouping
x32x2+4x8=x2(x2)+4(x2)=(x2)(x2+4)x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)
Group the terms in pairs and factor out the common binomial factor (x2)(x - 2).
3
Simplify the rational expression by canceling common factors
(x2)(x+2)(x2+4)(x2)(x2+4)=x+2\frac{(x - 2)(x + 2)(x^2 + 4)}{(x - 2)(x^2 + 4)} = x + 2
Cancel the non-zero common factor (x2)(x2+4)(x - 2)(x^2 + 4) present in both numerator and denominator.

Anahtar Kavram

Simplifying rational expressions using difference of squares and factoring by grouping
Soru 336Soru
For all real numbers xx, the algebraic expression
x627x4+3x2+9\frac{x^6 - 27}{x^4 + 3x^2 + 9}
can be simplified to the polynomial form ax2+bx+cax^2 + bx + c, where aa, bb, and cc are real constants. What is the value of a+b+ca + b + c?
Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The simplified expression is x23x^2 - 3, which corresponds to polynomial coefficients a=1a = 1, b=0b = 0, and c=3c = -3. The sum a+b+ca + b + c equals 2-2.
Factoring the numerator x627x^6 - 27 as a difference of cubes (x2)333(x^2)^3 - 3^3 produces (x23)(x4+3x2+9)(x^2 - 3)(x^4 + 3x^2 + 9). Canceling the non-zero factor (x4+3x2+9)(x^4 + 3x^2 + 9) from the numerator and denominator simplifies the expression to x23x^2 - 3. In standard form ax2+bx+cax^2 + bx + c, a=1a = 1, b=0b = 0, and c=3c = -3. Adding these coefficients gives 1+0+(3)=21 + 0 + (-3) = -2.

Adım Adım Çözüm

1
Factor the numerator using the difference of cubes formula
x627=(x2)333=(x23)(x4+3x2+9)x^6 - 27 = (x^2)^3 - 3^3 = (x^2 - 3)(x^4 + 3x^2 + 9)
The expression x627x^6 - 27 matches the pattern u3v3u^3 - v^3 with u=x2u = x^2 and v=3v = 3.
2
Simplify the rational expression by canceling the common quadratic-biquadratic factor
\frac{(x^2 - 3)(x^4 + 3x^2 + 9)}{x^4 + 3x^2 + 9} = x^2 - 3
Since x4+3x2+9>0x^4 + 3x^2 + 9 > 0 for all real numbers xx, the denominator is never zero, allowing direct cancellation of the common factor.
3
Match coefficients with ax2+bx+cax^2 + bx + c and calculate a+b+ca + b + c
a = 1, b = 0, c = -3 \implies a + b + c = 1 + 0 + (-3) = -2
Comparing x23=1x2+0x3x^2 - 3 = 1x^2 + 0x - 3 to ax2+bx+cax^2 + bx + c determines the values of constants aa, bb, and cc.

Anahtar Kavram

Difference of Cubes Factoring Identity
Tahmini Süre:1m 30s
Soru 337Soru
For all real numbers xx and yy such that xyx \neq -y and x2yx \neq -2y, which of the following expressions is equivalent to
x3+2x2yxy22y3x2+3xy+2y2?\frac{x^3 + 2x^2y - xy^2 - 2y^3}{x^2 + 3xy + 2y^2}?
Cevabı ve açıklamayı göster

Cevap: xyx - y

Cevap

xyx - y
Factoring the numerator by grouping gives x2(x+2y)y2(x+2y)=(x2y2)(x+2y)=(xy)(x+y)(x+2y)x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y). Factoring the denominator yields (x+y)(x+2y)(x + y)(x + 2y). Dividing the numerator by the denominator cancels the common factors (x+y)(x + y) and (x+2y)(x + 2y), leaving xyx - y.

Adım Adım Çözüm

1
Factor the numerator by grouping terms
x3+2x2yxy22y3=x2(x+2y)y2(x+2y)=(x2y2)(x+2y)=(xy)(x+y)(x+2y)x^3 + 2x^2y - xy^2 - 2y^3 = x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y)
Grouping pairs of terms allows factoring out common binomial factors.
2
Factor the quadratic denominator
x2+3xy+2y2=(x+y)(x+2y)x^2 + 3xy + 2y^2 = (x + y)(x + 2y)
Finding two terms whose sum is 3y3y and product is 2y22y^2 factors the quadratic in xx.
3
Simplify the rational expression by canceling non-zero common factors
(xy)(x+y)(x+2y)(x+y)(x+2y)=xy\frac{(x - y)(x + y)(x + 2y)}{(x + y)(x + 2y)} = x - y
Since xyx \neq -y and x2yx \neq -2y, the factors (x+y)(x + y) and (x+2y)(x + 2y) are non-zero and can be canceled.

Anahtar Kavram

Polynomial factoring by grouping, difference of squares, quadratic trinomial factoring, and simplifying rational algebraic expressions.
Tahmini Süre:1m 30s
Soru 338Soru

The length of a rectangular plot of land is 33 meters less than twice its width. If the area of the plot is 9090 square meters, what is the perimeter of the plot, in meters?

Cevabı ve açıklamayı göster

Cevap: 39

Cevap

The perimeter of the plot of land is 39 meters.
Setting the length to 2w32w - 3 gives an area equation of w(2w3)=90w(2w - 3) = 90, which expands and rearranges to 2w23w90=02w^2 - 3w - 90 = 0. Factoring this quadratic equation yields (2w15)(w+6)=0(2w - 15)(w + 6) = 0. Since width must be positive, w=7.5w = 7.5 meters, which means the length is 1212 meters. The perimeter is 2(12+7.5)=392(12 + 7.5) = 39 meters.

Adım Adım Çözüm

1
Express the length in terms of width and set up the area equation.
Let ww be the width of the rectangle. Length l=2w3l = 2w - 3. Area equation: w(2w3)=90w(2w - 3) = 90.
The area of a rectangle is equal to length multiplied by width.
2
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
2w23w90=02w^2 - 3w - 90 = 0
Distributing ww and subtracting 9090 from both sides puts the equation in standard quadratic form.
3
Factor the quadratic equation.
(2w15)(w+6)=0(2w - 15)(w + 6) = 0
Finding two numbers with a product of 2×(90)=1802 \times (-90) = -180 and a sum of 3-3 gives 15-15 and 1212.
4
Determine the valid physical dimensions.
w=7.5w = 7.5 meters and l=12l = 12 meters.
The root w=6w = -6 is discarded because physical length cannot be negative. Thus w=152=7.5w = \frac{15}{2} = 7.5 meters.
5
Calculate the perimeter.
Perimeter =2(l+w)=2(12+7.5)=39= 2(l + w) = 2(12 + 7.5) = 39 meters.
The perimeter of a rectangle is given by 2×(length+width)2 \times (\text{length} + \text{width}).

Anahtar Kavram

Solving quadratic word problems via factoring
Tahmini Süre:1m 30s
Soru 339Soru

For all real numbers xx such that x2x \neq -2, x2x \neq 2, and x4x \neq 4, which of the following expressions are equivalent to x416x2x34x24x+16\frac{x^4 - 16x^2}{x^3 - 4x^2 - 4x + 16}? Select all such expressions.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: x3+4x2x24\frac{x^3 + 4x^2}{x^2 - 4}; x2(x+4)(x2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}

Cevap

The equivalent expressions are x3+4x2x24\frac{x^3 + 4x^2}{x^2 - 4} and x2(x+4)(x2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}.
The given expression factors as x2(x4)(x+4)(x24)(x4)\frac{x^2(x - 4)(x + 4)}{(x^2 - 4)(x - 4)}. Canceling (x4)(x - 4) leaves x2(x+4)x24\frac{x^2(x + 4)}{x^2 - 4}. Expanding the numerator gives x3+4x2x24\frac{x^3 + 4x^2}{x^2 - 4}, and factoring the denominator further gives x2(x+4)(x2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}. Both represent valid equivalent forms of the expression.

Adım Adım Çözüm

1
Factor the numerator of the given algebraic expression.
x416x2=x2(x216)=x2(x4)(x+4)x^4 - 16x^2 = x^2(x^2 - 16) = x^2(x - 4)(x + 4)
Factor out the greatest common factor x2x^2, then apply the difference of squares identity a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b).
2
Factor the denominator by grouping terms.
x34x24x+16=x2(x4)4(x4)=(x24)(x4)=(x2)(x+2)(x4)x^3 - 4x^2 - 4x + 16 = x^2(x - 4) - 4(x - 4) = (x^2 - 4)(x - 4) = (x - 2)(x + 2)(x - 4)
Group the first two terms and last two terms, factor out common binomials, and expand the remaining difference of squares.
3
Simplify the full rational expression by canceling common factors.
x2(x4)(x+4)(x24)(x4)=x2(x+4)x24=x3+4x2x24\frac{x^2(x - 4)(x + 4)}{(x^2 - 4)(x - 4)} = \frac{x^2(x + 4)}{x^2 - 4} = \frac{x^3 + 4x^2}{x^2 - 4}
Since x4x \neq 4, cancel the common factor (x4)(x - 4) from both numerator and denominator.
4
Compare the simplified form to the given choices to identify all equivalent expressions.
Both x3+4x2x24\frac{x^3 + 4x^2}{x^2 - 4} and x2(x+4)(x2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)} match the simplified algebraic forms.
Expanding the numerator or factoring the denominator yields these two equivalent representations.

Anahtar Kavram

Simplifying rational algebraic expressions by polynomial factoring and term grouping.
Soru 340Soru

If r1r_1 and r2r_2 are the two distinct real solutions to the quadratic equation x24x21=0x^2 - 4x - 21 = 0, where r1>r2r_1 > r_2, what is the value of r1+2r2r_1 + 2r_2?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The correct value of r1+2r2r_1 + 2r_2 is 11.
Factoring the quadratic polynomial x24x21x^2 - 4x - 21 gives (x7)(x+3)=0(x - 7)(x + 3) = 0, which yields solutions x=7x = 7 and x=3x = -3. Applying the constraint r1>r2r_1 > r_2 identifies r1=7r_1 = 7 and r2=3r_2 = -3. Substituting these values into r1+2r2r_1 + 2r_2 yields 7+2(3)=76=17 + 2(-3) = 7 - 6 = 1.

Adım Adım Çözüm

1
Factor the quadratic equation
(x7)(x+3)=0(x - 7)(x + 3) = 0
Find two numbers that multiply to 21-21 and add up to 4-4, which are 7-7 and 33.
2
Solve for the roots of the equation
x=7x = 7 or x=3x = -3
Set each linear factor equal to zero: x7=0    x=7x - 7 = 0 \implies x = 7 and x+3=0    x=3x + 3 = 0 \implies x = -3.
3
Assign values to r1r_1 and r2r_2 based on the condition r1>r2r_1 > r_2
r1=7r_1 = 7 and r2=3r_2 = -3
Since 7>37 > -3, r1r_1 must be 77 and r2r_2 must be 3-3.
4
Evaluate the expression r1+2r2r_1 + 2r_2
7+2(3)=76=17 + 2(-3) = 7 - 6 = 1
Substitute r1=7r_1 = 7 and r2=3r_2 = -3 into the given expression.

Anahtar Kavram

Solving quadratic equations by factoring and evaluating expressions involving roots.
Tahmini Süre:1m 30s
ÖncekiSayfa 17 / 18Sonraki
Algebra Alıştırma Soruları — GRE General Test — Sayfa 17 | Examkin