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 :

(kx^2 + xy) (ky^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 :

(kx^2 + xy) (ky^2 + xy) = k25

kx^2 + xy + y^2 + xy = k25

kx^2 + 2xy + y^2 = k25

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

x+ 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 - d225y2 - 16

c2y2 - d225y2 - 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)+ 12a2 b2 + 22(b2)2

= (3a2)+ 12a2 b2 + (2b2)2

= (3a2)+ 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.

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