# SAT PRACTICE QUESTIONS ON SOLVING RADICAL EQUATIONS

To solve equations with square roots, we must know about the inverse operation of square root.

Inverse operation of square root is taking square.

Question 1 :

√ab = a-b

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

(a)  ab = a2-b  (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 = a- 2ab + b2

ab+2ab = a- 2ab + 2ab + b2

3ab = a+ 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

 x+3 = 11x = 11 - 3x = 8 x+3 = -11x = -11 - 3x = -14

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

√(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

2x= 50

Dividing by 2 on both sides, we get

x= 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

 ax + 3 = 6When x = -3a(-3) + 3 = 6-3a = 6 - 3-3a = 3a = -1 ax + 3 = -6When x = -3a(-3) + 3 = -6-3a = -6 - 3-3a = -9a = 3

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 = x- 2x + 1

x- 2x - 2x + 1 - 6 = 0

x- 5 = 0

x2 = 5

x = √5

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