PROBLEMS INVOLVING SQUARE ROOTS FOR SAT

Problem 1 :

In the equation y - y - √(4x² + 28) = 0, x > 0 and y = 8. What is the value of x ?

A)  3   B)  4    C)  5   D)  6

Solution :

y - √(4x² + 28) = 0

Applying y = 8, we get

8 - √4x² + 28 = 0

8 = √4x² + 28 

To remove square root on both sides, we take square.

4x² + 28 = 64

4x² = 64 - 28

4x² = 36

Dividing by 4 on both sides, we get

x² = 36/4

x² = 9

x = ± 3

So, option A is correct.

Problem 2 :

√(2k² + 17) - x = 0

If k > 0 and x = 7 in the equation above, what is the value of k ?

A)  2     B)  3      C)  4       D) 5 

Solution :

√(2k² + 17) - x = 0

When x = 7

√(2k² + 17) - 7 = 0

√(2k² + 17) = 7

Take square on both sides.

2k² + 17 = 7²

2k² + 17 = 49

2k² = 49 - 17

2k² = 32

k² = 32/2

k² = 16

k = ± 4

So, the value of k is 4, option C is correct.

Problem 3 :

√(x - a) = x - 4

If a = 2, what is the solution set of the equation ?

A)  {3, 6}     B)  {2}      C)  {3}       D) {6}

Solution :

√(x - a) = x - 4

Applying a = 2, we get

√(x - 2) = x - 4

Squaring on both sides

x - 2 = (x - 4)²

x - 2 = x² - 8x + 16

x² - 8x - x + 16 + 2 = 0

x² - 9x + 18 = 0

(x - 6)(x - 3) = 0

x = 6 and x = 3

So, option A is correct.

Problem 4 :

The expression

where a and b are positive constants and x > 1, find the value of a + b.

Solution :

a = 36, b = 73/8

a + b = 36 + (73/8)

= (288 + 73)/8

= 361/8

So, the value of a + b is 361/8.

Problem 5 :

√(k + 2) - x = 0

In the equation above k is a constant. If x = 9, what is the value of k ?

A)  1     B)  7      C)  16       D) 79

Solution :

√(k + 2) - x = 0

Applying the value of x, we get

√(k + 2) - 9 = 0

√(k + 2) = 9

Squaring on both sides

k + 2 = 81

Subtracting 2 on both sides,

x = 81 - 2

x = 79

So, option D is correct.

Problem 6 :

If √x + √9 = √64, find the value of x.

A)  √5      B)  5     C)  25    D)  55

Solution :

√x + √9 = √64

√x + 3 = 8

√x = 8 - 3

√x = 5

Squaring on both sides

x = 25

Problem 7 :

If √(2x + 6) + 4 = x + 3

What is the solution set of equation above ?

A)  {-1}   B)  {5}    C)  {-1, 5}     D)  {0, -1, 5}

Solution :

√(2x + 6) + 4 = x + 3

Subtracting 4,we get

√(2x + 6) = x + 3 - 4

√(2x + 6) = x - 1

Squaring on both sides

(2x + 6) = (x - 1)²

2x + 6 = x²- 2x + 1

x²- 2x - 2x + 1 - 6 = 0

x²- 4x - 5 = 0

(x - 5) (x + 1) = 0

x = 5 and x = -1

So, option C is correct.

Problem 8 :

If k = 1, which of the following is the solution set of

x - 7 = √(x - k) ?

A)  1    B)  5     C)  10    D) 5, 10

Solution :

x - 7 = √(x - k)

Applying the value of k, we get

x - 7 = √(x - 1)

Squaring on both sides, we get

(x - 7)² = x - 1

x² - 14x + 49 = x - 1

x² - 14x - x + 49 + 1 = 0

x² - 15x + 50 = 0

(x - 10) (x - 5) = 0

x = 10 and x = 5

Problem 9 :

If √49 = √y - √16,  what is the value of y ?

A)  121    B)  65     C)  11    D) √11

Solution :

√49 = √y - √16

7 = √y - 4

7 + 4 = √y

√y = 11

Squaring on both sides.

y = 11²

y = 121

Problem 10 :

√(c - 1) - t = 0

In the equation above, c is a constant. If t = 4, what is the value of c ?

A)  17     B)  15    C)  5     D)  3

Solution :

√(c - 1) - t = 0

Applying the value of t, we get

√(c - 1) - 4 = 0

√(c - 1) = 4

Squaring on both sides.

c - 1 = 16

Adding 1 on both sides

c = 16 + 1

c = 17

Problem 11 :

a - 6 = √(8a - 7) - 4

What is the solution set of the equation above ?

A)  0   B)  1    C)  11   D)  1, 11

Solution :

a - 6 = √(8a - 7) - 4

a - 6 + 4 = √(8a - 7)

a - 2 = √(8a - 7)

Squaring on both sides

(a - 2)² = 8a - 7

a² - 4a + 4 = 8a - 7

a² - 4a - 8a + 4 + 7 = 0

a² - 12a + 11 = 0

(a - 11) (a - 1) = 0

a = 1 and a = 11

So, the solution are 1 and 11.