SAT PRACTICE QUESTIONS WITH ALGEBRAIC IDENTITIES

Problem 1 :

If y > 0 and y² - 36 = 0, what is the value of y ?

Solution :

y² - 36 = 0

y² - 6² = 0

Using the algebraic identity a² - b², we expand it as (a - b) (a + b).

(y + 6)(y - 6) = 0

Equating each factor to 0, we get

y + 6 = 0 and y - 6 = 0

y = -6 and y = 6

Since y > 0, we choose y = 6.

Problem 2 :

Which expression is equivalent to 16x⁶ - 24x³y³ + 9y⁶ ?

Solution :

Given expression is 16x⁶ - 24x³y³ + 9y⁶

= 4²(x³)² - 24x³y³ + 3²(y³)²

= (4x³)² - 24x³y³ + (3y³)²

= (4x³)² - 2(4x³)(3y³) + (3y³)²

= (4x³ - 3y³)²

Problem 3 :

(k^{x^2 + xy}) (k^{y^2 + xy}) = k²⁵

In the equation above k > 1 and x = 3. What is the positive value of y ?

a)  1    b)  2    c)   4    d)  5

Solution :

(k^{x^2 + xy}) (k^{y^2 + xy}) = k²⁵

k^{x^2 + xy + y^2 + xy} = k²⁵

k^{x^2 + 2xy + y^2} = k²⁵

Since the bases area same on both sides of the equal sign, we equate the powers.

x² + 2xy + y² = 25

(x + y)² = 25

x + y = √25

x + y = 5 and x + y = -5

Applying x = 3 in x + y = 5, we get 3 + y = 5, y = 2

Applying x = 3 in x + y = -5, we get 3 + y = -5, y = -8

Problem 4 :

In the equation 3(x - 5)² + 7 = ax² + bx + c, a, b and c are constants. If the equation is true for all values of x, what is the value of c ?

Solution :

3(x - 5)² + 7 = ax² + bx + c

3(x² - 2x(5) + 5²) + 7 = ax² + bx + c

3(x² - 10x + 25) + 7 = ax² + bx + c

Distributing 3, we get

3x² - 30x + 75 + 7 = ax² + bx + c

3x² - 30x + 82 = ax² + bx + c

By comparing the corresponding terms, we get

a = 3, b = -30 and c = 82.

Problem 5 :

If (cy - d) (cy + d) = 25y² - 16

Which of the following could be the value of c in the equation above, where c and d are constants ?

a)  4     b)  5      c)  16     d)  25

Solution :

(cy - d) (cy + d) = 25y² - 16

(cy)² - d² = 25y² - 16

c²y² - d² = 25y² - 16

Comparing the corresponding terms, we get

c² = 25 and d² = 16

c = 5 and d = 4

Problem 6 :

If x² + y² = c  and -xy = b, which of the following is equivalent to c + 2b?

a)  (-2x - y)²    b)  (-x - y)²      c)  (x - y)²      d)   (x + y)² 

Solution :

x² + y² = c  -----(1)

-xy = b ------(2)

Multiplying by 2 on both sides, we get

-2xy = 2b

(1) + (2)

x² + y² - 2xy = c + 2b

(x - y)²  = c + 2b

So, option c is correct.

Problem 7 :

Which of the following is an equivalent form of 

(2.6a - 3.5)² - (7.3a² - 4.1)

a)  -2.1a² - 2.9      b)  -2.1a² + 11.1       c)  -0.54a² - 18.2a - 8.15

d)  -0.54a² + 18.2a + 16.35

Solution :

= (2.6a - 3.5)² - (7.3a² - 4.1)

= (2.6a)² - 2(2.6a) (3.5) + (3.5)² - 7.3a² + 4.1

= 6.76a² - 18.2a + (3.5)² - 7.3a² + 4.1

= 6.76a² - 18.2a + 12.25 - 7.3a² + 4.1

= -0.54a² - 18.2a + 16.35

So, option d is correct.

Problem 8 :

If x^a² / x^b² = x¹⁶, x > 1 and a + b = 2, what is the value of a - b?

a)  8     b)  14        c)  16        d)  18

Solution :

x^a² / x^b² = x¹⁶

x^(a² - b²) = x¹⁶

a² - b² = 16

(a + b)(a - b) = 16

Applying the value of a + b = 2, we get

2(a - b) = 16

a - b = 16/2

a - b = 8

So, option a is correct.

Problem 9 :

9a⁴ + 12a² b² + 4b⁴ 

Which of the following is equivalent to the expression shown above ?

a)  (3a² + 2b²)²            b)  (3a + 2b)⁴        c)  (9a² + 4b²)²          d)  (9a + 4b)⁴

Solution :

= 9a⁴ + 12a² b² + 4b⁴

= 3²(a²)² + 12a² b² + 2²(b²)²

= (3a²)² + 12a² b² + (2b²)²

= (3a²)² + 2(3a²)(2b²) + (2b²)²

= (3a² + 2b²)²

So, option a is correct.

Problem 10 :

(d − 30)(d + 30) − 7 = −7 What is a solution to the given equation?

Solution :

(d − 30)(d + 30) − 7 = −7

Adding 7 on both sides, we get

(d − 30)(d + 30) = −7 + 7

(d − 30)(d + 30) = 0

d - 30 = 0 and d + 30 = 0

d = 30 and d = -30

So, the solutions are -30 and 30.