SOLVING RADICAL EQUATIONS WITH TWO RADICALS

To remove radicals, for example

  • If we have square roots, to get rid of the square root take squares on both sides.
  • If we have cube roots, to get rid of the cube root take cubes  on both sides.

Some times we may have to use algebraic identities to find the expansion and solve it.

Solve the following radical equation

Problem 1 :

√(4x + 1) = √(x + 10)

Solution :

√(4x + 1) = √(x + 10)

Raise both sides to the power 2.

[√(4x + 1)]2 = (√x + 10)2

4x + 1 = x + 10

Subtracting x on both sides.

4x - x + 1 = 10

Subtracting 1 on both sides.

3x = 0

3x = 9

Dividing by 3 on both sides.

x = 3

Problem 2 :

√(3x – 3) - √(x + 12) = 0

Solution :

√(3x – 3) - √(x + 12) = 0

Add √(x + 12)

√(3x – 3) = √(x + 12)

Take squares on both sides.

3x - 3 = x + 12

Subtract x on both sides and add 3 on both sides.

3x - x = 12 + 3

2x = 15

Dividing by 2, we get

x = 15/2

Problem 3 :

2x 5 - 8x + 1 = 0

Solution :

∛(2x 5) - ∛(8x + 1) = 0

Add ∛(8x + 1)

(2x 5)3 = (8x + 1)3

2x 5 = 8x + 1

Subtract 8x and add 5 on both sides.

2x - 8x = 1 + 5

-6x = 6

Divide by -6 on both sides.

x = -6/6

x = -1

Problem 4 :

∛(x + 5) = 2∛(2x + 6)

Solution :

Given, x + 5 = 22x + 6

Raise both sides to the power 3.

(x + 5)3 = (2)3(2x + 6)3

x + 5 = (8)(2x + 6)

x + 5 = 16x + 48

x + 5 -16x 48 = 0

-15x 43 = 0

-15x = 43

x = -43/15

Problem 5 :

√(3x – 8) + 1 = √(x + 5)

Solution :

√(3x – 8) + 1 = √(x + 5) ----(1)

[√(3x – 8) + 1]2 = √(x + 5)

√(3x – 8)2 + 2 √(3x – 8) + (1)2 = √(x + 5)2

3x – 8 + 1 + 2√(3x – 8) = x + 5

3x - 7 + 2√(3x – 8) = x + 5

2x - 12 = -2√(3x – 8)

6 - x = √(3x – 8)

Take square on both sides.

(6 - x)2 = 3x - 8

36 - 2(6)x + x2 = 3x - 8

x2 - 12x + 36 = 3x - 8

Subtract 3x and add 8 on both sides.

x2 - 15x + 44 = 0

(x - 11)(x - 4) = 0

x = 11 and x = 4

By applying the value of x in (1), we get the solution as x = 4.

Problem 6 :

√(x + 2) = 2 - √x

Solution :

√(x + 2) = 2 - √x

√(x + 2) + √x = 2

Take squares on both sides.

(√(x + 2) + √x)2 = 22

x + 2 + x + 2√(x + 2)√x = 4

2x + 2 + 2√[x(x + 2)] = 4

Subtract 2x and 2 on both sides.

2√[x(x + 2)] = 4 - 2x - 2

2√[x2 + 2x)] = 2 - 2x

Dividing by 2, we get

√[x2 + 2x)] = 1 - x

Take square on both sides.

x2 + 2x = (1 - x)2

x2 + 2x = 1 + x2 - 2x

4x = 1

x = 1/4

Problem 7 :

√(2x + 4) = √(x + 2)

Solution :

√2x + 4 = √x + 2

√(2x + 4)2 = √(x + 2)2

2x + 4 = x + 2

Subtracting x and 4, we get

2x - x  = 2 - 4

x = -2

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More