Algebra

239 soru

Soru 1Soru →

A binary operation ⊗\otimes on the set of real numbers R\mathbb{R} is defined by a⊗b=a2+b2−aba \otimes b = a^2 + b^2 - ab. If x⊗3=19x \otimes 3 = 19 and x>0x > 0, find the value of xx.

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Cevap: 5

Cevap

The value of xx is 55.
Applying the operation rule gives x2+32−3x=19x^2 + 3^2 - 3x = 19, which simplifies to x2−3x−10=0x^2 - 3x - 10 = 0. Factoring this equation yields (x−5)(x+2)=0(x - 5)(x + 2) = 0. Since xx is constrained to be positive (x>0x > 0), the unique valid answer is 55.

Adım Adım Çözüm

1
Apply the definition of the binary operation to x⊗3x \otimes 3
x2+32−3(x)=x2−3x+9x^2 + 3^2 - 3(x) = x^2 - 3x + 9
Substitute a=xa = x and b=3b = 3 into a⊗b=a2+b2−aba \otimes b = a^2 + b^2 - ab.
2
Equate the result to 19 and rearrange into standard quadratic form
x2−3x−10=0x^2 - 3x - 10 = 0
Subtract 19 from both sides to set the quadratic equation to zero.
3
Factor the quadratic equation and solve for xx
(x−5)(x+2)=0  ⟹  x=5 or x=−2(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2
Find two numbers that multiply to −10-10 and add up to −3-3.
4
Apply the restriction x>0x > 0
x=5x = 5
Reject the negative solution x=−2x = -2 because xx must be strictly positive.

Anahtar Kavram

Evaluation of Binary Operations and Solving Quadratic Equations
Soru 2Soru →

The perimeter of a rectangular playfield is 28 m28\text{ m} and its area is 40 m240\text{ m}^2. What is the positive difference, in metres, between its length and width?

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Cevap: 6

Cevap

The positive difference between the length and width of the playfield is 6 metres.
Formulating the system gives x+y=14x + y = 14 and xy=40xy = 40. Substituting y=14−xy = 14 - x yields the quadratic equation x2−14x+40=0x^2 - 14x + 40 = 0, which factors into (x−10)(x−4)=0(x - 10)(x - 4) = 0. The dimensions are 10 m10\text{ m} and 4 m4\text{ m}, giving a positive difference of 10−4=6 m10 - 4 = 6\text{ m}.

Adım Adım Çözüm

1
Set up linear and quadratic equations for perimeter and area
x+y=14x + y = 14 and xy=40xy = 40
Perimeter formula is 2(x+y)=282(x + y) = 28 which simplifies to x+y=14x + y = 14, and area formula is xy=40xy = 40.
2
Substitute y=14−xy = 14 - x into the quadratic area equation
x(14−x)=40  ⟹  x2−14x+40=0x(14 - x) = 40 \implies x^2 - 14x + 40 = 0
Substitution reduces the simultaneous system to a single quadratic equation in terms of xx.
3
Solve the quadratic equation by factoring
(x−10)(x−4)=0  ⟹  x=10 or x=4(x - 10)(x - 4) = 0 \implies x = 10 \text{ or } x = 4
The roots of the equation give the dimensions of the rectangle.
4
Calculate the positive difference between the two dimensions
10 - 4 = 6
Subtract the smaller dimension from the larger dimension.

Anahtar Kavram

Solving word problems involving simultaneous linear and quadratic equations
Tahmini Süre:1m 30s
Soru 3Soru →

A quantity PP varies partially as xx and partially as the square of yy. When x=2x = 2 and y=3y = 3, P=24P = 24, and when x=5x = 5 and y=1y = 1, P=17P = 17. What is the value of PP when x=4x = 4 and y=3y = 3?

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Cevap: 30

Cevap

The value of PP is 30.
The relationship follows the partial variation formula P=k1x+k2y2P = k_1 x + k_2 y^2. Substituting the given conditions gives 2k1+9k2=242k_1 + 9k_2 = 24 and 5k1+k2=175k_1 + k_2 = 17. Solving these simultaneous equations yields k1=3k_1 = 3 and k2=2k_2 = 2. Evaluating P=3(4)+2(32)P = 3(4) + 2(3^2) produces 12+18=3012 + 18 = 30.

Adım Adım Çözüm

1
Set up the general formula for partial variation.
P=k1x+k2y2P = k_1 x + k_2 y^2, where k1k_1 and k2k_2 are constants.
Partial variation combines terms linearly with separate variation constants.
2
Substitute the given pairs of values to form simultaneous linear equations.
Equation (1): 2k1+9k2=242k_1 + 9k_2 = 24; Equation (2): 5k1+k2=175k_1 + k_2 = 17.
Plugging in (x=2,y=3,P=24)(x=2, y=3, P=24) and (x=5,y=1,P=17)(x=5, y=1, P=17) creates a system of equations in terms of k1k_1 and k2k_2.
3
Solve the simultaneous linear equations for k1k_1 and k2k_2.
From Equation (2), k2=17−5k1k_2 = 17 - 5k_1. Substitute into Equation (1): 2k1+9(17−5k1)=24  ⟹  −43k1=−129  ⟹  k1=32k_1 + 9(17 - 5k_1) = 24 \implies -43k_1 = -129 \implies k_1 = 3. Then k2=17−5(3)=2k_2 = 17 - 5(3) = 2.
Finding the specific values of the variation constants is required to complete the formula.
4
Calculate PP for x=4x = 4 and y=3y = 3 using the complete formula P=3x+2y2P = 3x + 2y^2.
P=3(4)+2(32)=12+2(9)=12+18=30P = 3(4) + 2(3^2) = 12 + 2(9) = 12 + 18 = 30.
Evaluating the relationship with the target parameters produces the final answer.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations
Tahmini Süre:2m 0s
Soru 4Soru →

Given the 3×33 \times 3 matrix A=(x213121x0)A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}, find the positive value of xx for which det⁡(A)=−2\det(A) = -2.

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Cevap: 2.5

Cevap

The positive value of xx is 2.5.
Expanding the determinant of matrix AA along the third row gives 1(4−1)−x(2x−3)=−2x2+3x+31(4 - 1) - x(2x - 3) = -2x^2 + 3x + 3. Setting this equal to −2-2 yields 2x2−3x−5=02x^2 - 3x - 5 = 0. Factoring gives (2x−5)(x+1)=0(2x - 5)(x + 1) = 0, yielding solutions x=2.5x = 2.5 and x=−1x = -1. Taking the positive value gives x=2.5x = 2.5.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA in terms of xx
det⁡(A)=−2x2+3x+3\det(A) = -2x^2 + 3x + 3
Expanding along the third row simplifies computation because of the zero entry.
2
Set the determinant expression equal to −2-2 and rearrange terms
2x2−3x−5=02x^2 - 3x - 5 = 0
Setting −2x2+3x+3=−2-2x^2 + 3x + 3 = -2 forms a standard quadratic equation.
3
Factorize the quadratic equation to find the roots
x=2.5x = 2.5 or x=−1x = -1
Factoring (2x−5)(x+1)=0(2x - 5)(x + 1) = 0 yields two real solutions.
4
Filter for the positive value requested in the stem
x=2.5x = 2.5
The question specifically requires the positive value of xx.

Anahtar Kavram

Determinant of a 3x3 Matrix and Quadratic Equation Solving
Tahmini Süre:2m 30s
Soru 5Soru →

Solve the simultaneous equations y=2x+1y = 2x + 1 and y=x2−2y = x^2 - 2. Which of the following represents the complete set of solution pairs (x,y)(x, y)?

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Cevap: (3,7)(3, 7) and (−1,−1)(-1, -1)

Cevap

The complete set of solution pairs (x,y)(x, y) is (3,7)(3, 7) and (−1,−1)(-1, -1).
Equating 2x+1=x2−22x + 1 = x^2 - 2 yields x2−2x−3=0x^2 - 2x - 3 = 0. Factoring gives (x−3)(x+1)=0(x - 3)(x + 1) = 0, leading to x=3x = 3 or x=−1x = -1. Substituting these xx-values into y=2x+1y = 2x + 1 gives y=7y = 7 for x=3x = 3, and y=−1y = -1 for x=−1x = -1. Thus, the solution pairs are (3,7)(3, 7) and (−1,−1)(-1, -1).

Adım Adım Çözüm

1
Equate the linear expression for yy to the quadratic expression for yy.
2x+1=x2−22x + 1 = x^2 - 2
Since both expressions equal yy, setting them equal eliminates yy.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−2x−3=0x^2 - 2x - 3 = 0
Subtract 2x2x and 11 from both sides.
3
Factor the quadratic equation to find the values of xx.
(x−3)(x+1)=0  ⟹  x=3 or x=−1(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1
Determine two numbers that multiply to −3-3 and add to −2-2.
4
Substitute each xx-value back into the linear equation y=2x+1y = 2x + 1 to find the corresponding yy-value.
For x=3x = 3, y=2(3)+1=7y = 2(3) + 1 = 7. For x=−1x = -1, y=2(−1)+1=−1y = 2(-1) + 1 = -1.
Calculate the exact coordinate pairs (x,y)(x, y) that satisfy both equations.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.
Soru 6Soru →

Find the number of integer values of xx that satisfy the inequality 3x2−10x−8≤03x^2 - 10x - 8 \leq 0.

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Cevap: 5

Cevap

The number of integer values of xx satisfying the inequality is 5.
Factoring 3x2−10x−8≤03x^2 - 10x - 8 \leq 0 gives (3x+2)(x−4)≤0(3x + 2)(x - 4) \leq 0. The region where the quadratic expression is non-positive lies between the roots x=−23x = -\frac{2}{3} and x=4x = 4, yielding −23≤x≤4-\frac{2}{3} \leq x \leq 4. The integers falling within this closed interval are 0,1,2,3,0, 1, 2, 3, and 44, giving a total of 5 integer solutions.

Adım Adım Çözüm

1
Factor the quadratic expression
(3x+2)(x−4)≤0(3x + 2)(x - 4) \leq 0
Factoring allows us to find the critical boundary values of the inequality.
2
Find the critical values (roots)
x=−23x = -\frac{2}{3} and x=4x = 4
Setting each linear factor to zero determines where the expression changes sign.
3
Determine the solution set interval
−23≤x≤4-\frac{2}{3} \leq x \leq 4
Since the coefficient of x2x^2 is positive, the quadratic curve is convex (U-shaped), so the expression is less than or equal to zero between the roots.
4
List and count the integer solutions
Integers: 0,1,2,3,40, 1, 2, 3, 4 (Total = 5)
The smallest integer greater than or equal to −23-\frac{2}{3} is 00, and the largest integer less than or equal to 44 is 44.

Anahtar Kavram

Solving quadratic inequalities and identifying integer solutions within a continuous range.
Soru 7Soru →

A binary operation ⋆\star on the set of real numbers is defined by a⋆b=3a+2b−1a \star b = 3a + 2b - 1. What is the value of xx such that 4⋆x=254 \star x = 25?

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Cevap: 7

Cevap

The value of xx is 77.
Applying the binary operation definition a⋆b=3a+2b−1a \star b = 3a + 2b - 1 with a=4a = 4 and b=xb = x gives 3(4)+2x−1=253(4) + 2x - 1 = 25. Simplifying gives 2x+11=252x + 11 = 25, leading to 2x=142x = 14 and x=7x = 7.

Adım Adım Çözüm

1
Substitute a=4a = 4 and b=xb = x into the definition of the binary operation a⋆b=3a+2b−1a \star b = 3a + 2b - 1.
4⋆x=3(4)+2x−14 \star x = 3(4) + 2x - 1
Applying the given rule for the binary operation.
2
Simplify the expression on the left-hand side.
4⋆x=12+2x−1=2x+114 \star x = 12 + 2x - 1 = 2x + 11
Performing basic arithmetic multiplication and addition of constants.
3
Set the simplified expression equal to 2525 and solve for xx.
2x+11=25  ⟹  2x=14  ⟹  x=72x + 11 = 25 \implies 2x = 14 \implies x = 7
Subtracting 1111 from both sides and dividing by 22.

Anahtar Kavram

Evaluation of non-commutative binary operations and solving algebraic equations involving operational rules.
Soru 8Soru →

If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the real solution pairs to the simultaneous equations x−2y=1x - 2y = 1 and x2−xy+y2=7x^2 - xy + y^2 = 7, what is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

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Cevap: −11-11

Cevap

The value of x1x2+y1y2x_1 x_2 + y_1 y_2 is −11-11.
Substituting x=2y+1x = 2y + 1 into x2−xy+y2=7x^2 - xy + y^2 = 7 gives 3y2+3y−6=03y^2 + 3y - 6 = 0, which reduces to y2+y−2=0y^2 + y - 2 = 0. The roots are y=1y = 1 and y=−2y = -2. Substituting back into the linear expression yields corresponding values x=3x = 3 and x=−3x = -3. The two solution pairs are (3,1)(3, 1) and (−3,−2)(-3, -2). Calculating x1x2+y1y2=(3)(−3)+(1)(−2)x_1 x_2 + y_1 y_2 = (3)(-3) + (1)(-2) yields −11-11.

Adım Adım Çözüm

1
Express xx in terms of yy from the linear equation.
x=2y+1x = 2y + 1
Isolating variable xx allows direct substitution into the quadratic equation.
2
Substitute x=2y+1x = 2y + 1 into the quadratic equation x2−xy+y2=7x^2 - xy + y^2 = 7.
(2y+1)2−(2y+1)y+y2=7  ⟹  3y2+3y−6=0(2y + 1)^2 - (2y + 1)y + y^2 = 7 \implies 3y^2 + 3y - 6 = 0
This simplifies the system to a single quadratic equation in yy.
3
Solve the quadratic equation 3y2+3y−6=03y^2 + 3y - 6 = 0.
y2+y−2=0  ⟹  (y+2)(y−1)=0y^2 + y - 2 = 0 \implies (y + 2)(y - 1) = 0, giving y1=1y_1 = 1 and y2=−2y_2 = -2
Factoring determines the two possible yy-coordinates.
4
Determine corresponding xx-coordinates for each yy-value.
For y1=1y_1 = 1, x1=2(1)+1=3  ⟹  (3,1)x_1 = 2(1) + 1 = 3 \implies (3, 1). For y2=−2y_2 = -2, x2=2(−2)+1=−3  ⟹  (−3,−2)x_2 = 2(-2) + 1 = -3 \implies (-3, -2).
Substituting each yy into x=2y+1x = 2y + 1 yields the complete solution pairs.
5
Compute x1x2+y1y2x_1 x_2 + y_1 y_2.
(3)(−3)+(1)(−2)=−9−2=−11(3)(-3) + (1)(-2) = -9 - 2 = -11
Evaluates the targeted algebraic expression.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.
Soru 9Soru →

If the matrix A=(3275)A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}, which matrix represents the inverse A−1A^{-1}?

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Cevap: (5−2−73)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

Cevap

(5−2−73)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
The correct inverse matrix is computed by evaluating the determinant det⁡(A)=3(5)−2(7)=1\det(A) = 3(5) - 2(7) = 1 and constructing the adjugate matrix by swapping the diagonal elements 33 and 55 while changing the signs of 22 and 77, resulting in (5−2−73)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA
det⁡(A)=(3)(5)−(2)(7)=15−14=1\det(A) = (3)(5) - (2)(7) = 15 - 14 = 1
The inverse requires dividing the adjugate matrix by the determinant of AA.
2
Find the adjugate of matrix AA
adj(A)=(5−2−73)\text{adj}(A) = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
For a 2x2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, swap the main diagonal elements (aa and dd) and change the signs of the off-diagonal elements (bb and cc).
3
Compute A−1=1det⁡(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A)
A−1=11(5−2−73)=(5−2−73)A^{-1} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
Multiply the adjugate matrix by the reciprocal of the determinant.

Anahtar Kavram

2x2 Matrix Inversion
Tahmini Süre:1m 0s
Soru 10Soru →

Solve for xx in the logarithmic equation log⁡2(x+2)+log⁡2(x−4)=4\log_2 (x + 2) + \log_2 (x - 4) = 4.

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Cevap: 6

Cevap

The value of xx is 6.
Using the product rule for logarithms, log⁡2(x+2)+log⁡2(x−4)=log⁡2[(x+2)(x−4)]=4\log_2 (x + 2) + \log_2 (x - 4) = \log_2 [(x + 2)(x - 4)] = 4. Converting to exponential form yields (x+2)(x−4)=24=16(x + 2)(x - 4) = 2^4 = 16. Expanding gives x2−2x−8=16x^2 - 2x - 8 = 16, which simplifies to x2−2x−24=0x^2 - 2x - 24 = 0. Factoring gives (x−6)(x+4)=0(x - 6)(x + 4) = 0. Since the logarithmic arguments must be positive (x>4x > 4), the negative root −4-4 is discarded, leaving x=6x = 6.

Adım Adım Çözüm

1
Apply the product rule of logarithms: log⁡bM+log⁡bN=log⁡b(M⋅N)\log_b M + \log_b N = \log_b (M \cdot N)
log⁡2[(x+2)(x−4)]=4\log_2 [(x + 2)(x - 4)] = 4
Logarithms with the same base being added combine by multiplying their arguments.
2
Convert the logarithmic equation to its exponential form
(x+2)(x−4)=24=16(x + 2)(x - 4) = 2^4 = 16
If log⁡bY=X\log_b Y = X, then Y=bXY = b^X.
3
Expand and rearrange into a standard quadratic equation
x2−2x−8=16  ⟹  x2−2x−24=0x^2 - 2x - 8 = 16 \implies x^2 - 2x - 24 = 0
Expanding (x+2)(x−4)(x+2)(x-4) yields x2−2x−8x^2 - 2x - 8, and subtracting 16 sets the equation to zero.
4
Factor the quadratic equation to find potential solutions
(x−6)(x+4)=0  ⟹  x=6 or x=−4(x - 6)(x + 4) = 0 \implies x = 6 \text{ or } x = -4
The roots of x2−2x−24=0x^2 - 2x - 24 = 0 are x=6x = 6 and x=−4x = -4.
5
Check the domain restrictions for logarithmic functions
x=6x = 6
The arguments of the original logarithms require x+2>0  ⟹  x>−2x + 2 > 0 \implies x > -2 and x−4>0  ⟹  x>4x - 4 > 0 \implies x > 4. Therefore, x=−4x = -4 is extraneous and x=6x = 6 is the only valid solution.

Anahtar Kavram

Solving logarithmic equations using product rule and domain constraints
Soru 11Soru →

If log⁡4x+log⁡2x=6\log_4 x + \log_2 x = 6, find the value of xx.

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Cevap: 16

Cevap

The value of xx is 16.
Using the change of base identity log⁡ba=log⁡calog⁡cb\log_b a = \frac{\log_c a}{\log_c b}, we convert log⁡4x\log_4 x into log⁡2xlog⁡24=log⁡2x2\frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}. Substituting this into the equation yields 32log⁡2x=6\frac{3}{2} \log_2 x = 6, which simplifies to log⁡2x=4\log_2 x = 4, leading directly to x=24=16x = 2^4 = 16.

Adım Adım Çözüm

1
Apply the change of base formula to log⁡4x\log_4 x
\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}
Converting all terms to a common base (base 2) simplifies addition of logarithmic terms.
2
Substitute the expression back into the original equation and collect like terms
\frac{1}{2}\log_2 x + \log_2 x = \frac{3}{2}\log_2 x = 6
Adding the coefficients of log⁡2x\log_2 x gives 32\frac{3}{2}.
3
Isolate log⁡2x\log_2 x
\log_2 x = 6 \cdot \frac{2}{3} = 4
Multiplying both sides by 23\frac{2}{3} isolates the logarithmic term.
4
Convert from logarithmic form to exponential form
x = 2^4 = 16
If log⁡ba=c\log_b a = c, then a=bca = b^c.

Anahtar Kavram

Change of base rule for logarithms
Soru 12Soru →

What is the positive integer solution to the logarithmic equation log⁡2x−3log⁡x2=2\log_2 x - 3\log_x 2 = 2?

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Cevap: 88

Cevap

The positive integer solution is 88.
Using the change of base identity log⁡x2=1log⁡2x\log_x 2 = \frac{1}{\log_2 x}, we rewrite the equation as u−3u=2u - \frac{3}{u} = 2 where u=log⁡2xu = \log_2 x. Rearranging yields the quadratic u2−2u−3=0u^2 - 2u - 3 = 0, which factors as (u−3)(u+1)=0(u - 3)(u + 1) = 0. This gives u=3u = 3 or u=−1u = -1. Converting back to x=2ux = 2^u, we get x=23=8x = 2^3 = 8 or x=2−1=12x = 2^{-1} = \frac{1}{2}. Since the question asks for the positive integer solution, the correct answer is 8.

Adım Adım Çözüm

1
Apply the change of base rule to express log⁡x2\log_x 2 in terms of base 2.
log⁡2x−3log⁡2x=2\log_2 x - \frac{3}{\log_2 x} = 2
The change of base identity states that log⁡ba=1log⁡ab\log_b a = \frac{1}{\log_a b}.
2
Substitute u=log⁡2xu = \log_2 x into the equation and clear the fraction.
u−3u=2  ⟹  u2−2u−3=0u - \frac{3}{u} = 2 \implies u^2 - 2u - 3 = 0
Multiplying through by uu (where u≠0u \neq 0) transforms the equation into standard quadratic form.
3
Factor the quadratic equation to solve for uu.
(u−3)(u+1)=0  ⟹  u=3 or u=−1(u - 3)(u + 1) = 0 \implies u = 3 \text{ or } u = -1
Finding the roots of the quadratic equation in terms of uu.
4
Convert back to xx using x=2ux = 2^u and select the positive integer solution.
For u=3u = 3: x=23=8x = 2^3 = 8. For u=−1u = -1: x=2−1=12x = 2^{-1} = \frac{1}{2}.
The question asks specifically for the positive integer solution, which is 8.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction
Soru 13Soru →

Find the value of xx that satisfies the exponential equation 27x−1=9x+13x−527^{x - 1} = \frac{9^{x + 1}}{3^{x - 5}}.

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Cevap: 5

Cevap

The value of xx is 5.
Rewriting the terms with a common base of 3 transforms the equation into 33x−3=3x+73^{3x-3} = 3^{x+7}. Equating exponents gives 3x−3=x+73x - 3 = x + 7, which yields x=5x = 5.

Adım Adım Çözüm

1
Convert all terms to base 3
27x−1=33x−327^{x-1} = 3^{3x-3} and 9x+1=32x+29^{x+1} = 3^{2x+2}
Laws of indices require identical bases to manipulate exponents.
2
Apply division rule of indices to the right-hand side
32x+23x−5=3x+7\frac{3^{2x+2}}{3^{x-5}} = 3^{x+7}
When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.
3
Equate the exponents and solve for xx
3x−3=x+7  ⟹  2x=10  ⟹  x=53x - 3 = x + 7 \implies 2x = 10 \implies x = 5
If af(x)=ag(x)a^f(x) = a^g(x) for a>0a > 0 and a≠1a \neq 1, then f(x)=g(x)f(x) = g(x).

Anahtar Kavram

Solving exponential equations using base reduction and exponent laws
Tahmini Süre:1m 30s
Soru 14Soru →

What is the square root of the surd expression 14+6514 + 6\sqrt{5}?

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Cevap: 3+53 + \sqrt{5}

Cevap

3+53 + \sqrt{5}
Expanding the square of 3+53 + \sqrt{5} yields (3)2+2(3)(5)+(5)2=9+65+5=14+65(3)^2 + 2(3)(\sqrt{5}) + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}, which accurately equals the original expression under the radical.

Adım Adım Çözüm

1
Set up the general form for the square root of a binomial surd
Let 14+65=a+b\sqrt{14 + 6\sqrt{5}} = \sqrt{a} + \sqrt{b}
The square root of a compound surd expression takes the form of a sum of radical terms
2
Square both sides of the equation
14+65=a+b+2ab14 + 6\sqrt{5} = a + b + 2\sqrt{ab}
Eliminate the outer radical to equate real and surd parts
3
Equate the rational parts and the surd parts
a+b=14a + b = 14 and 2ab=65  ⟹  ab=35=45  ⟹  ab=452\sqrt{ab} = 6\sqrt{5} \implies \sqrt{ab} = 3\sqrt{5} = \sqrt{45} \implies ab = 45
Match integer terms together and radical terms together
4
Solve for values of aa and bb
Two positive numbers with sum 1414 and product 4545 are 99 and 55, so a=9a = 9 and b=5b = 5
Determine the factors satisfying both equations
5
Substitute aa and bb into the radical expression
14+65=9+5=3+5\sqrt{14 + 6\sqrt{5}} = \sqrt{9} + \sqrt{5} = 3 + \sqrt{5}
Simplify 9\sqrt{9} to 33 to obtain the final simplified expression

Anahtar Kavram

Finding the square root of a surd expression by equating rational and radical parts
Soru 15Soru →

If 5+35−3−5−35+3=x15\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = x\sqrt{15}, what is the value of xx?

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Cevap: 2

Cevap

The value of xx is 2.
Rationalising both fractions yields 4+154 + \sqrt{15} and 4−154 - \sqrt{15}. Subtracting the second from the first gives (4+15)−(4−15)=215(4 + \sqrt{15}) - (4 - \sqrt{15}) = 2\sqrt{15}. Comparing 2152\sqrt{15} with x15x\sqrt{15} gives x=2x = 2.

Adım Adım Çözüm

1
Rationalise the denominator of the first fraction
\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = 4 + \sqrt{15}
Multiplying numerator and denominator by the conjugate of the denominator removes the surd from the denominator.
2
Rationalise the denominator of the second fraction
\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{5 - 2\sqrt{15} + 3}{5 - 3} = 4 - \sqrt{15}
Multiply by the conjugate (5−3)(\sqrt{5}-\sqrt{3}) to simplify the second surd term.
3
Subtract the simplified expressions
(4 + \sqrt{15}) - (4 - \sqrt{15}) = 4 - 4 + \sqrt{15} + \sqrt{15} = 2\sqrt{15}
Distribute the negative sign and combine like surd terms.
4
Solve for the unknown coefficient x
2\sqrt{15} = x\sqrt{15} \implies x = 2
Divide both sides of the equation by 15\sqrt{15} to isolate xx.

Anahtar Kavram

Binomial Surd Rationalisation and Simplification
Tahmini Süre:1m 30s
Soru 16Soru →

If x+4−x−1=1\sqrt{x + 4} - \sqrt{x - 1} = 1, what is the value of xx?

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Cevap: 55

Cevap

The value of xx is 55.
Isolating x+4\sqrt{x + 4} gives x+4=1+x−1\sqrt{x + 4} = 1 + \sqrt{x - 1}. Squaring both sides yields x+4=1+2x−1+x−1x + 4 = 1 + 2\sqrt{x - 1} + x - 1, which simplifies to 4=2x−14 = 2\sqrt{x - 1}. Dividing by 2 gives 2=x−12 = \sqrt{x - 1}. Squaring both sides once more gives 4=x−14 = x - 1, which leads to x=5x = 5.

Adım Adım Çözüm

1
Isolate one of the surd terms on the left side of the equation.
x+4=1+x−1\sqrt{x + 4} = 1 + \sqrt{x - 1}
Rearranging terms prevents dealing with complex cross-products when squaring.
2
Square both sides of the equation.
x+4=1+2x−1+(x−1)x + 4 = 1 + 2\sqrt{x - 1} + (x - 1)
Squaring eliminates the outer radical on the left side.
3
Simplify both sides and isolate the remaining radical term.
4=2x−1  ⟹  2=x−14 = 2\sqrt{x - 1} \implies 2 = \sqrt{x - 1}
Subtracting xx from both sides simplifies the algebraic expression.
4
Square both sides again to solve for xx.
4=x−1  ⟹  x=54 = x - 1 \implies x = 5
Squaring eliminates the remaining radical term.

Anahtar Kavram

Solving Surd Equations by Rational Elimination
Soru 17Soru →

Given the universal set U={x∈Z:1≤x≤15}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}, with subsets P={x∈U:x is a prime number}P = \{x \in \mathcal{U} : x \text{ is a prime number}\} and Q={x∈U:x is an odd number}Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}, what is the number of elements in the set (P∪Q)′(P \cup Q)'?

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Cevap: 6; six; 6 elements

Cevap

The number of elements in (P∪Q)′(P \cup Q)' is 6.
The set P∪QP \cup Q consists of all numbers from 1 to 15 that are either prime or odd: {1,2,3,5,7,9,11,13,15}\{1, 2, 3, 5, 7, 9, 11, 13, 15\}. The complement (P∪Q)′(P \cup Q)' relative to U\mathcal{U} contains all elements of U\mathcal{U} that are neither prime nor odd, which are the even composite numbers: {4,6,8,10,12,14}\{4, 6, 8, 10, 12, 14\}. Counting these elements gives 6.

Adım Adım Çözüm

1
List all elements of the universal set U\mathcal{U}, subset PP, and subset QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}, P={2,3,5,7,11,13}P = \{2, 3, 5, 7, 11, 13\}, and Q={1,3,5,7,9,11,13,15}Q = \{1, 3, 5, 7, 9, 11, 13, 15\}.
Listing the explicit elements allows precise execution of set union and complement operations.
2
Find the union P∪QP \cup Q.
P∪Q={1,2,3,5,7,9,11,13,15}P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}.
The union combines all unique elements that belong to either set PP, set QQ, or both.
3
Determine the complement set (P∪Q)′(P \cup Q)' relative to U\mathcal{U} and count its elements.
(P∪Q)′={4,6,8,10,12,14}(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}, which contains 6 elements.
The complement set consists of all elements in U\mathcal{U} that are not present in P∪QP \cup Q.

Anahtar Kavram

Complement of Set Union
Soru 18Soru →

When the polynomial P(x)=3x3−kx2+4x−7P(x) = 3x^3 - kx^2 + 4x - 7 is divided by x−2x - 2, the remainder is 99. What is the value of kk?

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Cevap: 4

Cevap

The value of kk is 44.
According to the Remainder Theorem, dividing P(x)P(x) by x−2x - 2 leaves a remainder of P(2)P(2). Evaluating P(2)=3(2)3−k(2)2+4(2)−7P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 gives 25−4k25 - 4k. Setting 25−4k=925 - 4k = 9 and solving yields k=4k = 4.

Adım Adım Çözüm

1
Apply the Remainder Theorem
The remainder when P(x)P(x) is divided by x−2x - 2 is P(2)P(2).
By the Remainder Theorem, dividing a polynomial P(x)P(x) by x−ax - a leaves a remainder equal to P(a)P(a).
2
Substitute x=2x = 2 into P(x)P(x)
P(2)=3(2)3−k(2)2+4(2)−7=25−4kP(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 = 25 - 4k
Evaluating the polynomial at x=2x = 2 expresses the remainder in terms of kk.
3
Equate P(2)P(2) to the given remainder and solve for kk
25−4k=9  ⟹  4k=16  ⟹  k=425 - 4k = 9 \implies 4k = 16 \implies k = 4
Setting the calculated expression equal to 99 forms a linear equation that yields k=4k = 4.

Anahtar Kavram

Polynomial Remainder Theorem
Tahmini Süre:1m 30s
Soru 19Soru →

What is the remainder when the polynomial P(x)=2x3−5x2+7x−3P(x) = 2x^3 - 5x^2 + 7x - 3 is divided by (2x−1)(2x - 1)?

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Cevap: −12-\frac{1}{2}

Cevap

The remainder is −12-\frac{1}{2}.
By the Remainder Theorem, when P(x)P(x) is divided by (ax−b)(ax - b), the remainder is P(ba)P\left(\frac{b}{a}\right). Setting 2x−1=02x - 1 = 0 yields x=12x = \frac{1}{2}. Evaluating P(12)=2(18)−5(14)+7(12)−3=14−54+72−3=−12P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + 7\left(\frac{1}{2}\right) - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -\frac{1}{2}.

Adım Adım Çözüm

1
Find the root of the linear divisor
2x−1=0  ⟹  x=122x - 1 = 0 \implies x = \frac{1}{2}
According to the Remainder Theorem, dividing P(x)P(x) by a linear divisor (ax−b)(ax - b) yields a remainder of P(ba)P\left(\frac{b}{a}\right).
2
Substitute x=12x = \frac{1}{2} into P(x)=2x3−5x2+7x−3P(x) = 2x^3 - 5x^2 + 7x - 3
P(12)=2(12)3−5(12)2+7(12)−3P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) - 3
Evaluating the polynomial at the root of the divisor determines the remainder.
3
Calculate the arithmetic value
P(12)=2(18)−5(14)+72−3=14−54+72−3=−1+72−3=−4+3.5=−0.5=−12P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + \frac{7}{2} - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -1 + \frac{7}{2} - 3 = -4 + 3.5 = -0.5 = -\frac{1}{2}
Simplifying the fractional terms gives the final remainder value.

Anahtar Kavram

Remainder Theorem for Linear Divisors (ax−b)(ax - b)
Soru 20Soru →

In a geometric progression of positive terms, the sum of the first two terms is 1212 and the sum of the third and fourth terms is 4848. What is the 6th6^{\text{th}} term of the progression?

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Cevap: 128

Cevap

The 6th term of the geometric progression is 128.
Dividing ar2(1+r)=48ar^2(1+r) = 48 by a(1+r)=12a(1+r) = 12 yields r2=4r^2 = 4, so r=2r = 2 for positive terms. Substituting r=2r = 2 into a(1+r)=12a(1+r) = 12 gives a=4a = 4. Using Tn=arn−1T_n = a r^{n-1} for n=6n=6, we get T6=4×25=128T_6 = 4 \times 2^5 = 128.

Adım Adım Çözüm

1
Set up algebraic equations for the given sums using first term aa and common ratio rr.
a(1+r)=12a(1+r) = 12 and ar2(1+r)=48ar^2(1+r) = 48
The terms of a geometric progression are given by Tn=arn−1T_n = a r^{n-1}.
2
Divide the equation for the third and fourth terms by the equation for the first and second terms.
r2=4  ⟹  r=2r^2 = 4 \implies r = 2
Dividing eliminates aa and (1+r)(1+r), giving r2=4r^2 = 4. Since terms are positive, r>0r > 0.
3
Substitute r=2r = 2 into a(1+r)=12a(1+r) = 12 to solve for aa.
a=4a = 4
3a=123a = 12 leads directly to a=4a = 4.
4
Evaluate the 6th term using the formula T6=ar5T_6 = a r^{5}.
T6=4×25=128T_6 = 4 \times 2^5 = 128
Applying the general term formula Tn=arn−1T_n = a r^{n-1} with n=6n=6.

Anahtar Kavram

Geometric Progression term relations and finding the common ratio from consecutive term pairs
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