SAT PRACTICE QUESTIONS ON SOLVING RADICAL EQUATIONS

Question 1 :

√ab = a-b

If a > 0 and b > 0, the equation above is equivalent to which of the following ?

(a)  ab = a²-b²   (b) ab = a²+b²

(c)  2ab - a²-b²  (d) 3ab = a²+b²

Solution :

√ab = a-b

For removing square root, we can take squares on both sides.

ab = (a-b)²

ab = a² - 2ab + b²

Adding 2ab on both sides.

ab+2ab = a² - 2ab + 2ab + b²

3ab = a² + b²

Question 2 :

Solve for x, when x > 0

(x + 3)² = 121

Solution :

(x + 3)² = 121

By taking square roots on both sides, we get

√(x + 3)² = √121

(x + 3) = ±11

Question 3 :

If √x + √9 = √64, what is the value of x ?

(a)  √5  (b)  5  (c)  25  (d)  55

Solution :

√x + √9 = √64

√9 = 3 and √64 = 8

√x + 3 = 8

√x  = 8 - 3

√x  = 5

Taking squares on both sides.

x  = 5²

x = 25

Question 4 :

√(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

By applying the value of x in the given equation.

√(2k²+17) - 7 = 0

Add 7 on both sides.

√(2k²+17) = 7

Taking squares on both sides, we get

2k²+17 = 49

Subtract 17 on both sides, we get

2k² = 49-17

2k² = 32

Dividing by 2 on both sides,

k² = 16

k = ±4

Question 5 :

If √x+ √y = 4√y, where x > 0 and y > 0, what is x in terms of y ?

(a)  16y   (b)  9y   (c)  6y   (d)  4y

Solution :

√x + √y = 4√y

Subtract √y on both sides.

√x  = 4√y - √y

√x  = 3√y

Take squares on both sides.

x = 9y

Question 6 :

√(2x²-14)/a = 3

If x > 0 and a = 2 in the equation above, what is the value of x ?

(a)  4   (b)  5   (c)  6   (d)  7

Solution :

√(2x²-14)/a = 3

By applying the value of a, we get

√(2x²-14)/2 = 3

√(2x²-14) = 6

By taking squares on both sides, we get

(2x²-14) = 6²

(2x²-14) = 36

2x² = 36+14

2x² = 50

Dividing by 2 on both sides, we get

x² = 25

x = 5

Question 7 :

If x > 0 and 9x² = 40, which of the following is equivalent to the value of x ?

(a)  (40/9)²   (b)  √40/9   (c)  √9/40   (d)  √(40/9)

Solution :

9x² = 40

Divide by 9 on both sides.

x² = 40/9

Taking square root on both sides.

x = √(40/9)

Question 8 :

If m = 1/√n, where m > 0 and n > 0, what is n in terms of m ?

(a)  n = 1/√m   (b)  n = 1/m   (c)  n = 1/m²   (d)  n = m²

Solution :

m = 1/√n

Taking squares on both sides, we get

m² = (1/√n)²

m² = 1/n

n = 1/m²

Question 9 :

If √(4+√x) = 1 + √3, what is the value of x ?

(a)  0   (b)  2   (c)  6   (d)  12

Solution :

√(4+√x) = 1 + √3

[√(4+√x)]² = (1 + √3)²

4+√x = 1 + 2√3 + 3

4+√x = 4 + 2√3

Subtracting 4 on both sides.

√x = 2√3

x = (2√3)²

x = 4(3)

x = 12

Question 10 :

If 3√x³ = √72, what is the value of x ?

Solution :

3√x³ = √72

Take squares on both sides.

9x³ = 72

Divide by 9 on both sides.

x³ = 72/9

x³ = 8

x³ = 2³

x = 2

Question 11 :

In the equation (ax + 3)² = 36 , a is a constant. If x = −3 is one solution to the equation, what is a possible value of a ?

a) −11   b) −5    c) −1    d) 0

Solution :

(ax + 3)² = 36 

To remove the square, we take square roots on both sides.

ax + 3 = √36

ax + 3 = ±6

Question 12 :

√(2x+6) + 4 = x + 3

 What is the solution set of the equation above?

a) {−1}   b) {5}   c) {−1, 5}    d) {0, −1, 5}

Solution :

√(2x + 6) + 4 = x + 3

Subtract 4 from both sides.

√(2x + 6) = x + 3 - 4

√(2x + 6) = x - 1

Taking squares on both sides.

(2x + 6) = (x - 1)²

2x + 6 = x² - 2x + 1

x² - 2x - 2x + 1 - 6 = 0

x² - 5 = 0

x² = 5

x = √5