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Question 341Question →
For all real numbers xx and yy such that x≠yx \neq y, which of the following expressions is equivalent to x3−x2y−xy2+y3x−y−(x−y)2\frac{x^3 - x^2y - xy^2 + y^3}{x - y} - (x - y)^2?
Show answer & explanation

Answer: 2xy−2y22xy - 2y^2

Answer

The equivalent expression is 2xy−2y22xy - 2y^2.
Factoring the numerator by grouping gives x2(x−y)−y2(x−y)=(x2−y2)(x−y)x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y). Dividing by (x−y)(x - y) yields x2−y2x^2 - y^2. Expanding (x−y)2(x - y)^2 gives x2−2xy+y2x^2 - 2xy + y^2. Subtracting this expansion from x2−y2x^2 - y^2 results in (x2−y2)−(x2−2xy+y2)=2xy−2y2(x^2 - y^2) - (x^2 - 2xy + y^2) = 2xy - 2y^2.

Step-by-Step Solution

1
Factor the numerator of the rational fraction by grouping terms
x3−x2y−xy2+y3=x2(x−y)−y2(x−y)=(x2−y2)(x−y)=(x+y)(x−y)2x^3 - x^2y - xy^2 + y^3 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x + y)(x - y)^2
Grouping terms allows factoring out common binomial components.
2
Simplify the fraction by dividing by (x−y)(x - y)
\frac{(x + y)(x - y)^2}{x - y} = (x + y)(x - y) = x^2 - y^2
Since x≠yx \neq y, x−y≠0x - y \neq 0, so common factors can be canceled.
3
Expand the subtracted term (x−y)2(x - y)^2
(x - y)^2 = x^2 - 2xy + y^2
Applying the standard perfect square trinomial expansion formula.
4
Subtract the expanded expression from the simplified fraction result
(x^2 - y^2) - (x^2 - 2xy + y^2) = x^2 - y^2 - x^2 + 2xy - y^2 = 2xy - 2y^2
Distribute the negative sign across all terms inside the parentheses and combine like terms.

Key Concept

Simplifying algebraic expressions using factoring by grouping, difference of squares, and expanding binomial squares.
Question 342Question →

If mm and nn are the two distinct real solutions to the quadratic equation 2x2−8x+c=02x^2 - 8x + c = 0, where cc is a constant, and m2+n2=12m^2 + n^2 = 12, what is the value of cc?

Show answer & explanation

Answer: 4

Answer

The value of cc is 44.
By Vieta's formulas for the quadratic equation 2x2−8x+c=02x^2 - 8x + c = 0, the sum of the roots is m+n=4m + n = 4 and the product of the roots is mn=c/2mn = c/2. Squaring the sum gives (m+n)2=m2+2mn+n2=16(m + n)^2 = m^2 + 2mn + n^2 = 16. Substituting m2+n2=12m^2 + n^2 = 12 gives 12+2(c/2)=1612 + 2(c/2) = 16, which simplifies to 12+c=1612 + c = 16, so c=4c = 4.

Step-by-Step Solution

1
Apply Vieta's formulas to determine the sum and product of the roots in terms of the given quadratic coefficients.
For 2x2−8x+c=02x^2 - 8x + c = 0, the sum of roots is m+n=−−82=4m + n = -\frac{-8}{2} = 4, and the product of roots is m⋅n=c2m \cdot n = \frac{c}{2}.
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is −b/a-b/a and the product of roots is c/ac/a.
2
Relate m2+n2m^2 + n^2 to the sum and product of roots using the identity (m+n)2=m2+n2+2mn(m + n)^2 = m^2 + n^2 + 2mn.
m2+n2=(m+n)2−2mnm^2 + n^2 = (m + n)^2 - 2mn
Rewriting the sum of squares in terms of (m+n)(m+n) and mnmn allows substitution of the known values.
3
Substitute m+n=4m + n = 4, mn=c2mn = \frac{c}{2}, and m2+n2=12m^2 + n^2 = 12 into the equation and solve for cc.
12=(4)2−2(c2)  ⟹  12=16−c  ⟹  c=412 = (4)^2 - 2\left(\frac{c}{2}\right) \implies 12 = 16 - c \implies c = 4.
Direct algebraic simplification yields the constant cc.

Key Concept

Relationship between roots and coefficients of a quadratic equation (Vieta's Formulas) and algebraic identities.
Question 343Question →
For all real numbers xx such that x≠−3x \neq -3, x≠0x \neq 0, and x≠3x \neq 3, which of the following expressions is equivalent to
x4−81x3+3x2÷x2+9x2\frac{x^4 - 81}{x^3 + 3x^2} \div \frac{x^2 + 9}{x^2}?
Show answer & explanation

Answer: x−3x - 3

Answer

x−3x - 3
Factoring the numerator x4−81x^4 - 81 gives (x−3)(x+3)(x2+9)(x - 3)(x + 3)(x^2 + 9) and factoring the denominator x3+3x2x^3 + 3x^2 gives x2(x+3)x^2(x + 3). Multiplying by the reciprocal x2x2+9\frac{x^2}{x^2 + 9} allows the factors (x+3)(x + 3), (x2+9)(x^2 + 9), and x2x^2 to cancel completely, leaving x−3x - 3.

Step-by-Step Solution

1
Rewrite division of rational expressions as multiplication by the reciprocal
x4−81x3+3x2⋅x2x2+9\frac{x^4 - 81}{x^3 + 3x^2} \cdot \frac{x^2}{x^2 + 9}
Dividing by a fraction is equivalent to multiplying by its reciprocal.
2
Factor the numerator x4−81x^4 - 81 using the difference of squares formula twice
x4−81=(x2−9)(x2+9)=(x−3)(x+3)(x2+9)x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)
Recognize x4−81x^4 - 81 as (x2)2−92(x^2)^2 - 9^2 and then factor x2−9x^2 - 9 further into (x−3)(x+3)(x - 3)(x + 3).
3
Factor out the greatest common factor from the denominator x3+3x2x^3 + 3x^2
x3+3x2=x2(x+3)x^3 + 3x^2 = x^2(x + 3)
Extract x2x^2 from both terms of the binomial.
4
Substitute the factored expressions into the product and cancel common factors
(x−3)(x+3)(x2+9)x2(x+3)⋅x2x2+9=x−3\frac{(x - 3)(x + 3)(x^2 + 9)}{x^2(x + 3)} \cdot \frac{x^2}{x^2 + 9} = x - 3
The common terms (x+3)(x + 3), (x2+9)(x^2 + 9), and x2x^2 cancel out completely from the numerator and denominator for all x≠−3,0,3x \neq -3, 0, 3.

Key Concept

Factoring polynomial expressions using difference of squares and common terms to simplify rational expressions.
Estimated Time:1m 30s
Question 344Question →

For what positive value of the constant kk does the quadratic equation x2−2kx+3k+10=0x^2 - 2kx + 3k + 10 = 0 have exactly one real solution?

Show answer & explanation

Answer: 5

Answer

The positive value of kk is 5.
A quadratic equation has exactly one real solution when its discriminant b2−4acb^2 - 4ac equals 00. For x2−2kx+(3k+10)=0x^2 - 2kx + (3k + 10) = 0, substituting a=1a = 1, b=−2kb = -2k, and c=3k+10c = 3k + 10 gives (−2k)2−4(1)(3k+10)=0(-2k)^2 - 4(1)(3k + 10) = 0, which simplifies to 4k2−12k−40=04k^2 - 12k - 40 = 0. Dividing the entire equation by 44 gives k2−3k−10=0k^2 - 3k - 10 = 0. Factoring this quadratic yields (k−5)(k+2)=0(k - 5)(k + 2) = 0, giving solutions k=5k = 5 and k=−2k = -2. Since the question requires the positive value of kk, the correct answer is 55.

Step-by-Step Solution

1
Identify the condition for a quadratic equation to have exactly one real solution.
The discriminant must equal zero (b2−4ac=0b^2 - 4ac = 0).
A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has a single repeated real root if and only if its discriminant is zero.
2
Set up the discriminant equation using the coefficients from x2−2kx+(3k+10)=0x^2 - 2kx + (3k + 10) = 0.
Discriminant =(−2k)2−4(1)(3k+10)=4k2−12k−40=0= (-2k)^2 - 4(1)(3k + 10) = 4k^2 - 12k - 40 = 0.
Here a=1a = 1, b=−2kb = -2k, and c=3k+10c = 3k + 10.
3
Divide by 4 and solve the quadratic in terms of kk by factoring.
k2−3k−10=0⇒(k−5)(k+2)=0k^2 - 3k - 10 = 0 \Rightarrow (k - 5)(k + 2) = 0, so k=5k = 5 or k=−2k = -2.
Factoring the simplified quadratic yields the two potential values of kk.
4
Choose the value of kk that satisfies the constraint in the stem.
k=5k = 5.
The question asks specifically for the positive value of kk, so k=−2k = -2 is rejected.

Key Concept

Quadratic Discriminant and Factoring
Estimated Time:1m 30s
Question 345Question →

For all real numbers xx such that x≠−2x \neq -2, which of the following expressions are equivalent to x6−64x3+8\frac{x^6 - 64}{x^3 + 8}? Select all such expressions.

Select all that apply

Show answer & explanation

Answer: x3−8x^3 - 8; (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4)

Answer

The expressions equivalent to the given rational expression are x3−8x^3 - 8 and (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4).
Factoring the numerator x6−64x^6 - 64 as a difference of squares yields (x3−8)(x3+8)(x^3 - 8)(x^3 + 8). Dividing by the denominator (x3+8)(x^3 + 8) leaves x3−8x^3 - 8. Further factoring x3−8x^3 - 8 as a difference of cubes gives (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4). Both x3−8x^3 - 8 and (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4) are valid equivalent expressions.

Step-by-Step Solution

1
Factor the numerator using the difference of squares identity
x6−64=(x3)2−82=(x3−8)(x3+8)x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8)
The expression x6−64x^6 - 64 is a difference of two squares.
2
Simplify the fraction by canceling the common non-zero term (x3+8)(x^3 + 8)
\frac{(x^3 - 8)(x^3 + 8)}{x^3 + 8} = x^3 - 8
Since x≠−2x \neq -2, x3+8≠0x^3 + 8 \neq 0, so (x3+8)(x^3 + 8) can be canceled from numerator and denominator.
3
Factor x3−8x^3 - 8 using the difference of cubes identity
x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)
Applying a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a=xa = x and b=2b = 2.

Key Concept

Simplifying rational expressions by factoring polynomial numerators and denominators using algebraic identities such as difference of squares and difference of cubes.
Question 346Question →

If xx is a real number that satisfies the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0, which of the following values could be the value of xx? Select all such values.

Select all that apply

Show answer & explanation

Answer: −3-3; −2-2; 33

Answer

The correct values are −3-3, −2-2, and 33.
Factoring the equation yields (x2−4)(x2−9)=(x−2)(x+2)(x−3)(x+3)=0(x^2 - 4)(x^2 - 9) = (x - 2)(x + 2)(x - 3)(x + 3) = 0, giving the four distinct real solutions x=−3,−2,2,3x = -3, -2, 2, 3. The options corresponding to −3-3, −2-2, and 33 represent valid values of xx.

Step-by-Step Solution

1
Substitute a variable u=x2u = x^2 to express the equation in quadratic form.
u2−13u+36=0u^2 - 13u + 36 = 0
This reduces the fourth-degree polynomial into a standard quadratic equation.
2
Factor the quadratic equation to solve for uu.
(u−4)(u−9)=0  ⟹  u=4 or u=9(u - 4)(u - 9) = 0 \implies u = 4 \text{ or } u = 9
Factoring determines the values of x2x^2.
3
Substitute back x2=ux^2 = u and solve for xx by taking both positive and negative square roots.
x2=4  ⟹  x=±2x^2 = 4 \implies x = \pm 2, and x2=9  ⟹  x=±3x^2 = 9 \implies x = \pm 3
Each positive value of uu yields two real solutions for xx.

Key Concept

Solving quadratic-form equations by factoring and taking positive and negative square roots.
Question 347Question →
For all real numbers xx such that x≠−3x \neq -3, x≠0x \neq 0, and x≠3x \neq 3, which of the following expressions is equivalent to x−2−9−1x−1+3−1⋅3x2+9xx−3?\frac{x^{-2} - 9^{-1}}{x^{-1} + 3^{-1}} \cdot \frac{3x^2 + 9x}{x - 3}?
Show answer & explanation

Answer: −x−3-x - 3

Answer

−x−3-x - 3
The numerator of the first factor is a difference of squares (x−1−3−1)(x−1+3−1)(x^{-1} - 3^{-1})(x^{-1} + 3^{-1}). Dividing by x−1+3−1x^{-1} + 3^{-1} leaves x−1−3−1=1x−13=3−x3xx^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}. Factoring 3x3x out of the numerator of the second expression gives 3x(x+3)3x(x + 3). Multiplying these terms yields 3−x3x⋅3x(x+3)x−3\frac{3 - x}{3x} \cdot \frac{3x(x + 3)}{x - 3}. Since 3−x=−(x−3)3 - x = -(x - 3), the binomials (3−x)(3 - x) and (x−3)(x - 3) cancel to −1-1, and 3x3x cancels out completely, resulting in −(x+3)=−x−3-(x + 3) = -x - 3.

Step-by-Step Solution

1
Factor the numerator of the first expression as a difference of squares.
x−2−9−1=(x−1)2−(3−1)2=(x−1−3−1)(x−1+3−1)x^{-2} - 9^{-1} = (x^{-1})^2 - (3^{-1})^2 = (x^{-1} - 3^{-1})(x^{-1} + 3^{-1})
Expressing negative exponents as squares allows cancellation with the denominator.
2
Divide by (x−1+3−1)(x^{-1} + 3^{-1}) and convert negative exponents into a single rational term.
\frac{(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})}{x^{-1} + 3^{-1}} = x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}
Finding a common denominator simplifies the first factor.
3
Factor the numerator of the second expression 3x2+9x3x^2 + 9x.
3x^2 + 9x = 3x(x + 3)
Factoring out the greatest common factor 3x3x enables further simplification.
4
Multiply the simplified expressions and reduce.
\left(\frac{3 - x}{3x}\right) \cdot \left(\frac{3x(x + 3)}{x - 3}\right) = \frac{-(x - 3)}{3x} \cdot \frac{3x(x + 3)}{x - 3} = -(x + 3) = -x - 3
Canceling 3x3x and noting that (3−x)/(x−3)=−1(3 - x)/(x - 3) = -1 leaves the linear expression −x−3-x - 3.

Key Concept

Simplifying algebraic expressions containing negative exponents, rational fractions, and opposite-sign binomial factors
Question 348Question →

When the algebraic expression 4x3−16x2x2+4x\frac{4x^3 - 16x}{2x^2 + 4x} is simplified for all x≠0x \neq 0 and x≠−2x \neq -2, it reduces to the linear polynomial ax+bax + b, where aa and bb are constants. What is the value of a+ba + b?

Show answer & explanation

Answer: -2

Answer

The correct numerical answer is -2.
Factoring the numerator yields 4x(x−2)(x+2)4x(x - 2)(x + 2) and factoring the denominator yields 2x(x+2)2x(x + 2). Canceling the common factors 2x2x and (x+2)(x + 2) leaves 2(x−2)=2x−42(x - 2) = 2x - 4. Comparing 2x−42x - 4 to ax+bax + b gives a=2a = 2 and b=−4b = -4. Summing these values gives a+b=2+(−4)=−2a + b = 2 + (-4) = -2.

Step-by-Step Solution

1
Factor out the greatest common factor and apply the difference of squares formula to the numerator.
4x3−16x=4x(x2−4)=4x(x−2)(x+2)4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)
Fully factoring the numerator allows identification of all linear factors.
2
Factor out the greatest common factor from the denominator.
2x2+4x=2x(x+2)2x^2 + 4x = 2x(x + 2)
Extracting 2x2x reveals the common terms shared with the numerator.
3
Divide the numerator by the denominator by canceling identical non-zero factors 2x2x and (x+2)(x + 2).
4x(x−2)(x+2)2x(x+2)=42(x−2)=2(x−2)=2x−4\frac{4x(x - 2)(x + 2)}{2x(x + 2)} = \frac{4}{2}(x - 2) = 2(x - 2) = 2x - 4
Simplifying rational expressions requires canceling common factors present in both numerator and denominator.
4
Compare the simplified expression 2x−42x - 4 to the form ax+bax + b to find aa and bb, then compute their sum.
a=2a = 2 and b=−4  ⟹  a+b=2+(−4)=−2b = -4 \implies a + b = 2 + (-4) = -2
Matching corresponding terms identifies the values of the target constants.

Key Concept

Simplifying rational expressions by factoring out common terms and applying the difference of squares identity.
Question 349Question →

If the quadratic equation 2x2−kx+18=02x^2 - kx + 18 = 0 has two positive real roots x1x_1 and x2x_2 such that x1=4x2x_1 = 4x_2, what is the value of the constant kk?

Show answer & explanation

Answer: 1515

Answer

1515
By Vieta's formulas for ax2+bx+c=0ax^2 + bx + c = 0, the product of the roots is x1x2=ca=182=9x_1 x_2 = \frac{c}{a} = \frac{18}{2} = 9. Given x1=4x2x_1 = 4x_2, substituting gives 4x22=94x_2^2 = 9, which yields x2=32x_2 = \frac{3}{2} since roots are positive. Then x1=6x_1 = 6. The sum of the roots is x1+x2=152x_1 + x_2 = \frac{15}{2}. By Vieta's formulas, x1+x2=−ba=k2x_1 + x_2 = -\frac{b}{a} = \frac{k}{2}. Setting k2=152\frac{k}{2} = \frac{15}{2} gives k=15k = 15.

Step-by-Step Solution

1
Apply Vieta's formula for the product of the roots
x1⋅x2=182=9x_1 \cdot x_2 = \frac{18}{2} = 9
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the product of the roots is ca\frac{c}{a}.
2
Substitute the given relationship x1=4x2x_1 = 4x_2 into the product equation
(4x2)(x2)=9  ⟹  4x22=9  ⟹  x22=94  ⟹  x2=32(4x_2)(x_2) = 9 \implies 4x_2^2 = 9 \implies x_2^2 = \frac{9}{4} \implies x_2 = \frac{3}{2}
Since the roots are positive, we take the positive square root of 94\frac{9}{4}.
3
Calculate the larger root x1x_1
x1=4(32)=6x_1 = 4 \left(\frac{3}{2}\right) = 6
Using the relationship x1=4x2x_1 = 4x_2.
4
Apply Vieta's formula for the sum of the roots to solve for kk
x1+x2=6+32=152=k2  ⟹  k=15x_1 + x_2 = 6 + \frac{3}{2} = \frac{15}{2} = \frac{k}{2} \implies k = 15
For ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −ba=k2-\frac{b}{a} = \frac{k}{2}.

Key Concept

Relating roots of a quadratic equation to its coefficients using Vieta's formulas and factoring relationships.
Question 350Question →

If aa and bb are the two real solutions to the quadratic equation 2x2−11x+12=02x^2 - 11x + 12 = 0, such that a>ba > b, what is the value of a−2ba - 2b?

Show answer & explanation

Answer: 1

Answer

The value of a−2ba - 2b is 1.
Factoring 2x2−11x+12=02x^2 - 11x + 12 = 0 yields (2x−3)(x−4)=0(2x - 3)(x - 4) = 0, giving solutions x=1.5x = 1.5 and x=4x = 4. Given that a>ba > b, we must set a=4a = 4 and b=1.5b = 1.5. Substituting these values into a−2ba - 2b gives 4−2(1.5)=14 - 2(1.5) = 1.

Step-by-Step Solution

1
Factor the quadratic equation 2x2−11x+12=02x^2 - 11x + 12 = 0
(2x−3)(x−4)=0(2x - 3)(x - 4) = 0
Find two linear factors whose product expands to 2x2−11x+122x^2 - 11x + 12.
2
Find the roots of the equation
x=32=1.5x = \frac{3}{2} = 1.5 and x=4x = 4
Apply the zero product property: 2x−3=0⇒x=1.52x - 3 = 0 \Rightarrow x = 1.5 and x−4=0⇒x=4x - 4 = 0 \Rightarrow x = 4.
3
Assign values to aa and bb based on the inequality a>ba > b
a=4a = 4 and b=1.5b = 1.5
Since 4>1.54 > 1.5, aa must be 4 and bb must be 1.5.
4
Evaluate the targeted expression a−2ba - 2b
4−2(1.5)=14 - 2(1.5) = 1
Substitute a=4a = 4 and b=1.5b = 1.5 into a−2ba - 2b.

Key Concept

Factoring Quadratic Equations
Question 351Question →

For all real numbers xx and yy such that y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, which of the following expressions is equivalent to (x+y)3−(x−y)32y(3x2+y2)\frac{(x + y)^3 - (x - y)^3}{2y(3x^2 + y^2)}?

Show answer & explanation

Answer: 11

Answer

11
Expanding the numerator yields (x3+3x2y+3xy2+y3)−(x3−3x2y+3xy2−y3)=6x2y+2y3=2y(3x2+y2)(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3 = 2y(3x^2 + y^2). Since the numerator and denominator are identical non-zero expressions, the quotient simplifies to 1.

Step-by-Step Solution

1
Expand (x+y)3(x + y)^3 and (x−y)3(x - y)^3 using the binomial theorem.
(x+y)3=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 and (x−y)3=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
Expanding the binomial expressions allows combination of like terms in the numerator.
2
Subtract (x−y)3(x - y)^3 from (x+y)3(x + y)^3.
(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3
The terms x3x^3 and 3xy23xy^2 subtract to zero, while the remaining terms double.
3
Factor out common terms from the simplified numerator.
6x2y+2y3=2y(3x2+y2)6x^2y + 2y^3 = 2y(3x^2 + y^2)
Factoring out 2y2y reveals a factor identical to the denominator.
4
Divide the numerator by the denominator 2y(3x2+y2)2y(3x^2 + y^2).
2y(3x2+y2)2y(3x2+y2)=1\frac{2y(3x^2 + y^2)}{2y(3x^2 + y^2)} = 1
Since y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, identical non-zero factors cancel out.

Key Concept

Binomial expansion of cubic expressions and factoring algebraic expressions.
Estimated Time:1m 30s
Question 352Question →

If mm and nn are positive integers such that the quadratic equation x2−mx+n=0x^2 - mx + n = 0 has two roots that are distinct prime numbers, and the discriminant of the equation is equal to 11, what is the value of m+nm + n?

Show answer & explanation

Answer: 11

Answer

11
For the quadratic equation x2−mx+n=0x^2 - mx + n = 0 with roots p1p_1 and p2p_2, the discriminant is D=(p1−p2)2D = (p_1 - p_2)^2. Given D=1D = 1, the difference between the two prime roots is ∣p1−p2∣=1|p_1 - p_2| = 1. The only prime numbers with a difference of 1 are 2 and 3. Thus, m=2+3=5m = 2 + 3 = 5 and n=2×3=6n = 2 \times 3 = 6, giving m+n=11m + n = 11.

Step-by-Step Solution

1
Express the discriminant of the quadratic equation in terms of its roots.
If p1p_1 and p2p_2 are the roots of x2−mx+n=0x^2 - mx + n = 0, Vieta's formulas give p1+p2=mp_1 + p_2 = m and p1p2=np_1 p_2 = n. The discriminant is D=m2−4n=(p1+p2)2−4p1p2=(p1−p2)2D = m^2 - 4n = (p_1 + p_2)^2 - 4p_1 p_2 = (p_1 - p_2)^2.
Relating the discriminant directly to the difference of the roots simplifies the constraint.
2
Determine the roots using the given discriminant value.
Since D=1D = 1, we have (p1−p2)2=1(p_1 - p_2)^2 = 1, which implies ∣p1−p2∣=1|p_1 - p_2| = 1.
Taking the square root of both sides indicates the two roots differ by 1.
3
Identify the prime numbers that satisfy this condition.
The only pair of prime numbers that differ by 1 is 2 and 3, because 2 is the only even prime and all other primes are odd.
Consecutive integers that are both prime must be 2 and 3.
4
Calculate mm, nn, and their sum m+nm + n.
m=2+3=5m = 2 + 3 = 5 and n=2×3=6n = 2 \times 3 = 6. Therefore, m+n=5+6=11m + n = 5 + 6 = 11.
Substitute the root values into the sum and product formulas.

Key Concept

Quadratic discriminant and root relationships (Vieta's Formulas)
Estimated Time:1m 30s
Question 353Question →

If xx and yy are non-zero real numbers such that x≠yx \neq y, x2−y2x−y=12\frac{x^2 - y^2}{x - y} = 12, and x2y−xy2xy=4\frac{x^2y - xy^2}{xy} = 4, what is the value of x2+y2x^2 + y^2?

Show answer & explanation

Answer: 80

Answer

The value of x2+y2x^2 + y^2 is 80.
Simplifying the first equation by factoring the difference of squares (x−y)(x+y)x−y=12\frac{(x-y)(x+y)}{x-y} = 12 gives x+y=12x + y = 12. Simplifying the second equation by factoring out xyxy yields xy(x−y)xy=4\frac{xy(x-y)}{xy} = 4, which gives x−y=4x - y = 4. Solving this system yields x=8x = 8 and y=4y = 4. Squaring and summing these values gives 82+42=64+16=808^2 + 4^2 = 64 + 16 = 80. Alternatively, using the algebraic identity (x+y)2+(x−y)22=122+422=144+162=80\frac{(x+y)^2 + (x-y)^2}{2} = \frac{12^2 + 4^2}{2} = \frac{144 + 16}{2} = 80 directly yields the correct answer.

Step-by-Step Solution

1
Simplify the first given algebraic expression using the difference of squares identity.
x+y=12x + y = 12
Factoring x2−y2x^2 - y^2 gives (x−y)(x+y)(x - y)(x + y). Since x≠yx \neq y, we can cancel the non-zero common factor (x−y)(x - y) from the numerator and denominator.
2
Simplify the second given algebraic expression by factoring out the greatest common factor.
x−y=4x - y = 4
Factoring xyxy from x2y−xy2x^2y - xy^2 yields xy(x−y)xy(x - y). Since x,y≠0x, y \neq 0, we cancel xyxy from the numerator and denominator.
3
Solve for the individual values of xx and yy.
x=8x = 8 and y=4y = 4
Adding (x+y=12)(x + y = 12) and (x−y=4)(x - y = 4) gives 2x=162x = 16, so x=8x = 8. Subtracting the equations gives 2y=82y = 8, so y=4y = 4.
4
Calculate the target expression x2+y2x^2 + y^2.
80
x2+y2=82+42=64+16=80x^2 + y^2 = 8^2 + 4^2 = 64 + 16 = 80.

Key Concept

Factoring Algebraic Expressions (Difference of Squares and GCF Extraction)
Question 354Question →

If the quadratic equation 3x2−12x+c=03x^2 - 12x + c = 0 has two real roots, r1r_1 and r2r_2, such that r12+r22=10r_1^2 + r_2^2 = 10, what is the value of the constant cc?

Show answer & explanation

Answer: 9

Answer

The value of the constant cc is 9.
The correct answer is 9. By Vieta's formulas, the sum of the roots is r1+r2=−−123=4r_1 + r_2 = -\frac{-12}{3} = 4 and the product of the roots is r1r2=c3r_1 r_2 = \frac{c}{3}. Using the identity r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2, we substitute the given values to get 10=16−2c310 = 16 - \frac{2c}{3}, which simplifies to 2c3=6\frac{2c}{3} = 6, giving c=9c = 9.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of the roots in terms of equation coefficients.
r1+r2=−−123=4r_1 + r_2 = -\frac{-12}{3} = 4 and r1r2=c3r_1 r_2 = \frac{c}{3}.
For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −ba-\frac{b}{a} and the product of the roots is ca\frac{c}{a}.
2
Relate the sum of squares r12+r22r_1^2 + r_2^2 to (r1+r2)(r_1 + r_2) and r1r2r_1 r_2.
r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2.
Expanding (r1+r2)2=r12+2r1r2+r22(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2 and rearranging gives the identity for the sum of squares.
3
Substitute the known values into the identity and solve for cc.
10=42−2(c3)  ⟹  10=16−2c3  ⟹  2c3=6  ⟹  c=910 = 4^2 - 2\left(\frac{c}{3}\right) \implies 10 = 16 - \frac{2c}{3} \implies \frac{2c}{3} = 6 \implies c = 9.
Substituting r12+r22=10r_1^2 + r_2^2 = 10 and r1+r2=4r_1 + r_2 = 4 isolates the single variable cc.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Alternative Method

Alternatively, factor 3(x2−4x+3)=03(x^2 - 4x + 3) = 0 directly once r1r_1 and r2r_2 are identified. Since r1+r2=4r_1 + r_2 = 4 and r12+r22=10r_1^2 + r_2^2 = 10, solving the system of equations for r1r_1 and r2r_2 yields roots of 1 and 3. The product of these roots is (1)(3)=3(1)(3) = 3, so c3=3\frac{c}{3} = 3, which gives c=9c = 9.
Estimated Time:1m 30s
Question 355Question →

For all real numbers xx and yy such that x≠2yx \neq 2y and x≠−2yx \neq -2y, which of the following expressions is equivalent to x3+2x2y−4xy2−8y3x2−4y2\frac{x^3 + 2x^2y - 4xy^2 - 8y^3}{x^2 - 4y^2}?

Show answer & explanation

Answer: x+2yx + 2y

Answer

The simplified expression is x+2yx + 2y.
Grouping terms in the numerator gives x2(x+2y)−4y2(x+2y)=(x2−4y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y). Dividing this by the denominator (x2−4y2)(x^2 - 4y^2) cancels out the identical non-zero factor (x2−4y2)(x^2 - 4y^2), leaving the linear binomial x+2yx + 2y.

Step-by-Step Solution

1
Group the four terms in the numerator in pairs to factor by grouping
x3+2x2y−4xy2−8y3=x2(x+2y)−4y2(x+2y)x^3 + 2x^2y - 4xy^2 - 8y^3 = x^2(x + 2y) - 4y^2(x + 2y)
Grouping the first two terms and the last two terms allows factoring out x2x^2 and −4y2-4y^2 respectively.
2
Factor out the common binomial factor (x+2y)(x + 2y) from the numerator
x2(x+2y)−4y2(x+2y)=(x2−4y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y)
Both terms share the common factor (x+2y)(x + 2y).
3
Divide the factored numerator by the denominator (x2−4y2)(x^2 - 4y^2)
(x2−4y2)(x+2y)x2−4y2=x+2y\frac{(x^2 - 4y^2)(x + 2y)}{x^2 - 4y^2} = x + 2y
Since x≠±2yx \neq \pm 2y, x2−4y2≠0x^2 - 4y^2 \neq 0, allowing the common polynomial factor (x2−4y2)(x^2 - 4y^2) to be canceled.

Key Concept

Factoring Four-Term Polynomials by Grouping and Rational Expression Simplification
Question 356Question →

If xx is a negative real number such that 2x2+5x−12=02x^2 + 5x - 12 = 0, what is the value of (2x−3)2(2x - 3)^2?

Show answer & explanation

Answer: 121

Answer

121
Factoring 2x2+5x−12=02x^2 + 5x - 12 = 0 gives (2x−3)(x+4)=0(2x - 3)(x + 4) = 0, which yields the solutions x=32x = \frac{3}{2} and x=−4x = -4. Because the problem specifies that xx is negative, xx must equal −4-4. Substituting −4-4 for xx in the target expression yields (2(−4)−3)2=(−11)2=121(2(-4) - 3)^2 = (-11)^2 = 121.

Step-by-Step Solution

1
Factor the given quadratic equation to find its roots.
2x2+5x−12=(2x−3)(x+4)=02x^2 + 5x - 12 = (2x - 3)(x + 4) = 0, giving solutions x=32x = \frac{3}{2} and x=−4x = -4.
Factoring isolates the linear terms to solve for the values of xx.
2
Apply the given constraint to select the appropriate root.
Since xx is specified as a negative real number, x=−4x = -4.
The root x=32x = \frac{3}{2} is positive and violates the problem constraint.
3
Substitute the valid root into the target expression (2x−3)2(2x - 3)^2.
(2(−4)−3)2=(−8−3)2=(−11)2=121(2(-4) - 3)^2 = (-8 - 3)^2 = (-11)^2 = 121.
Evaluating the expression with x=−4x = -4 gives the final requested numerical value.

Key Concept

Solving quadratic equations by factoring and evaluating algebraic expressions under given sign constraints.
Estimated Time:1m 30s
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