# ALGEBRAIC IDENTITIES PROBLEMS ON SAT

Problem 1 :

(3x + 2y)2

In the expression above can be written as ax2 + bxy + cy2, where a, b and c are constants, where a + b + c ?

Solution :

(3x + 2y)2

Using the algebraic identity,

(a + b)2 = a2 + 2ab + b2

Here a = 3x and b = 2y

(3x + 2y)2 = (3x)2 + 2(3x)(2y) + (2y)2

= 9x2 + 12xy + 4y2

By comparing with ax2 + bxy + cy2, we get

a = 9, b = 12 and c = 4

a + b + c = 9 + 12 + 4

= 25

Problem 2 :

x2 + kx + 9 = (x + a)2

In the equation above, k and a are positive constants. If the equation is true for all values of  x, what is the value of k ?

a)  0      b) 3     c)  6       d)  9

Solution :

x2 + kx + 9 = (x + a)2

Applying the algebraic identity, (a + b)= a2 + 2ab + b2

x2 + kx + 9 = x2 + 2ax + a2

Comparing the coefficients, we get

a2 = 9 and k = 2a

a = 3

Applying the value of a, we get

k = 2(3)

k = 6

So, the value of k is 6.

Problem 3 :

If (x + 3) (x - 3) = 91, what is the value of x2 ?

Solution :

(x + 3) (x - 3) = 91

Using the algebraic identity,

(a + b)(a - b) = a2 - b2

x2 - 32 = 91

x2 - 9 = 91

x2 = 91 + 9

x2 = 100

So, the value of x2 is 100.

Problem 4 :

If (c + d) = -5 and (c - d) = -12, then what is the value of c2 - d2?

Solution :

This problem can be done in two ways,

1) using the method of elimination

2) using algebraic identity

By using algebraic identity, we can do the problem simply

(a + b)(a - b) = a2 - b2

(c + d) (c - d) = -5 (-12)

c2 - d= 60

So, the value of c2 - dis 60.

Problem 5 :

If (mx + c) (nx + 3) = 12x2 + 5x - 3 for all values of x, where m, n and c are constants, what is the value of m + n ?

a) 7      b)  8     c)  12       d)  13

Solution :

(mx + c) (nx + 3) = 12x2 + 5x - 3

mnx2 + 3mx + cnx + 3c = 12x2 + 5x - 3

mnx2 + (3m + cn)x  + 3c = 12x2 + 5x - 3

Comparing the coefficients of x2, x and constant.

mn = 12 -----(1)

3m + cn = 5 -----(2)

3c = -3

c = -1

Applying the value of c in (2), we get

3m + (-1)n = 5

3m - n = 5

From (1),

m = 12/n

3(12/n) - n = 5

(36/n) - n = 5

(36 - n2)/n = 5

36 - n2 = 5n

n2 + 5n - 36 = 0

(n + 9)(n - 4) = 0

Equating each factor to 0, we get

n + 9 = 0 and n - 4 = 0

n = -9 and n = 4

When n = -9, m = 12/(-9) ==>  -4/3

When n = 4, m = 12/4 ==>  3

m + n = (-4/3) + (-9) ==> (-4-27)/3 ==> -31/3

m + n = 4 + 3 ==> 7

So, the value of m + n is 7, option a is correct.

Problem 6 :

In the equation above, a, b and c are constants. If the equation is true for all values of x, what is the value of a + b + c?

Solution :

Comparing the corresponding terms, we get

a = 30, b = 5 and c = 20

a + b + c = 30 + 5 + 20

= 55

So, the value of a + b + c is 55.

Problem 7 :

If (x + y)2 - (x - y)2 = 60 and x and y are positive integers, which of the following could be the value of x + y ?

a)  6      b)  8       c)  10    d)  12

Solution :

(x + y)2 - (x - y)2 = 60

x2 + 2xy + y2 - (x2 - 2xy + y2) = 60

x2 + 2xy + y2 - x2 + 2xy - y2 = 60

4xy = 60

When x = 3 and y = 5, then 4xy = 4(3)(5) ==> 60

Then the values of x and y are 3 and y respectively.

x + y = 3 + 5 ==> 8

So, the value of x + y is 8.

Problem 8 :

(a + b)2 - (a - b)2

The expression above equivalent to which of following ?

a) 2ab    b)  4ab     c)  4ab + 2b2      d)  2a2 + 2b2

Solution :

(a + b)2 - (a - b)2

Using the algebraic identities,

= (a + b)2 - (a - b)2

= (a2 + 2ab + b2) - (a2 - 2ab + b2)

= a2 + 2ab + b2 - a2 + 2ab - b2

= 2ab + 2ab

= 4ab

So, the answer is option b.

Problem 9 :

(x - c)2 = x + 3

If c = 3, what is the solution set of the equation above ?

a)  {1}    b)  {6}    c)  {1, 6}     d)  {-3, 1, 6}

Solution :

(x - c)2 = x + 3

When c = 3, we get

(x - 3)2 = x + 3

x2 - 2(x) (3) + 32 = x + 3

x2 - 6x + 9 = x + 3

x2 - 6x - x + 9 - 3 = 0

x2 - 7x + 6 = 0

Solving this quadratic equation by factoring,

(x - 6)(x - 1) = 0

Equation each factor to 0, we get

x - 6 = 0 and x - 1 = 0

x = 6 and x = 1

So, the values of x are {1, 6}, option c is correct.

Problem 10 :

x2 - y2 = 48

x + y = 12

If (x, y) is the solution to the system of equations above, what is the value of xy?

a)  28    b)  32    c)  45    d)  64

Solution :

x2 - y2 = 48 ----(1)

x + y = 12 ----(2)

Using algebraic identity,

x2 - y2 = 48

(x + y) (x - y) = 48

Applying the value of (2), we get

12 (x - y) = 48

x - y = 48/12

x - y = 4 ----(3)

(2) + (3)

x + y + x - y = 12 + 4

2y = 16

y = 8

Applying the value of y in (2), we get

x + 8 = 12

x = 12 - 8

x = 4

x y = 4(8) ==> 32

So, the value of xy is 32.

## Recent Articles

1. ### Finding Range of Values Inequality Problems

May 21, 24 08:51 PM

Finding Range of Values Inequality Problems

2. ### Solving Two Step Inequality Word Problems

May 21, 24 08:51 AM

Solving Two Step Inequality Word Problems