PROBLEMS USING ALGEBRAIC IDENTITIES IN SAT

Problem 1 :

Which of the following is equivalent to 

(1/m)² - 2 (1/m) (1/n) + (1/n)²

(a) (1/√m - 1/√n)⁴     (b) (1/m  - 1/n)²

(c)  1/(m-n)²     (d)  2/(m² - mn +n²)

Solution :

= (1/m)² - 2 (1/m) (1/n) + (1/n)²

The given expansion is in the form of

(a - b)² = a² - 2ab + b²

Here a = 1/m and b = 1/n

So, (1/m - 1/n)²

Problem 2 :

(3x + 2y)²

If the expression above can be written as

ax² + bxy + cy²

 where a, b and c are constants, what is the value a + b +c ?

Solution :

(3x + 2y)² 

Here a = 3x and b = 2y

= (3x)² + 2(3y)(2y) + (2y)²

= 3²x² + 2(3y)(2y) + (2²y)²

= 9x² + 12xy + 4y²

By comparing the corresponding terms, we get

9x² + 12xy + 4y² = ax² + bxy + cy²

a = 9, b = 12 and c = 4

a + b + c = 9 + 12 + 4

= 25

Problem 3 :

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 ?

Solution :

Using the algebraic identity

(a + b)² = a² + 2ab + b²

(x + a)² = x² + 2ax + a²  ------(1)

By applying (1) in the given question, we get

x² + kx + 9 = x² + 2ax + a²

Comparing the corresponding terms, we get

2a = k and a² = 9

a² = 3², then a = 3

2(3) = k

k = 6

Problem 4 :

2x (x - y) (x + y)

Which of the following is equivalent to the expression above ?

(a) 4x³ - 2xy²     (b)  2x³ + 2xy²

(c)  2x³ - 2xy²     (d)  2x³ - 4xy + 2xy²

Solution :

2x (x - y) (x + y)

In the given question (x - y) (x + y) looks like (a - b) (a + b)

(a - b) (a + b) = a² - b²

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

= 2x (x² - y²)

By distributing 2x, we get

= 2x³ - 2xy²

Problem 5 :

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

Solution :

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

Using algebraic identity (a + b)(a - b) = a² - b²

x² - 3² = 91

x² - 9 = 91

Add 9 on both sides, we get

x² = 91 + 9

x² = 100

So, the value of x² is 100.

Problem 6 :

The expression

(x + 1)/(x + 2) - (x - 2)/(x - 1)

is equivalent to which of the following ?

(a)  -5/(x + 2) (x - 1)    (b)  1/(x + 2) (x - 1)

(c)  3/(x + 2) (x - 1)    (b)  (2x²+3)/(x + 2) (x - 1)

Solution :

To combine the rational expression, let us take the least common multiple.

= [(x + 1)⋅(x - 1)/(x + 2)⋅(x -1)]  - [(x - 2)⋅(x+2)/(x - 1)⋅(x+2)]

Using the algebraic identity,

(a + b) (a- b) = a² - b²

we get

(x + 1)⋅(x - 1) = x² - 1²

(x - 2)⋅(x+2) = x² - 2²

= ((x² - 1) - (x² - 4)) / (x - 1)⋅(x + 2)

= (-1 + 4) / (x - 1)(x + 2)

= 3/(x - 1)(x + 2)

Problem 7 :

Which of the following is equivalent to

(m + n - 1) (m + n +1) ?

 (a) m² + 2mn + n² - 1      (b) m² - 2mn + n² - 1

(c) m² - n² - 1  (d) m² - 2mn + n² - 1

Solution :

(m + n - 1) (m + n +1)

Let a = m + n

= (a - 1) (a + 1)

= a² - 1 -----(1)

Applying the value of a in (1), we get

= (m + n)² - 1

= m² + n² - 1

Problem 8 :

(a + b)² - (a - b)²

The expression above is equivalent 

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

Solution :

(a + b)² - (a - b)²

= a² + 2ab + b² + a² - 2ab + b²

= 2a² + 2b²

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

By applying the value of c = 3 in the equation, we get

(x - 3)² = x + 3

x² - 2(x) (3) + 3² = x + 3

Subtracting x and 3, we get

x² - 6x + 9 - x - 3 = 0

x² - 7x + 6 = 0

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

x = 1 and x = 6

So, the value of x is {1, 6}.

Problem 10 :

The following is equivalent form of 

(1.5x - 2.4)² - (5.2x² - 6.4) ?

Solution :

Using the algebraic identity,

(a - b)² = a² - 2ab + b²

(1.5x - 2.4)² = (1.5x)² - 2(1.5x) (2.4) + (2.4)²

= 2.25x² - 7.2x + 5.76

(1.5x - 2.4)² - (5.2x² - 6.4) 

= 2.25x² - 7.2x + 5.76 - 5.2x² + 6.4

= 2.25x² - 5.2x² - 7.2x + 5.76 + 6.4

= -2.95x² - 7.2x + 12.16