﻿ Solve the cube root equation \sqrt{x+109} - \sqrt{x-109} = 2

#### Question

Solve the cube root equation

\sqrt{x+109} - \sqrt{x-109} = 2

Collected in the board: Cube root

Steven Zheng posted 4 months ago

Let

a = \sqrt{x+109}
(1)
b = \sqrt{x-109}
(2)

Then

a - b = 2
(3)

Difference of a and b raised to the power of 3 gives

a^3 - b^3 = 218
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 218
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 109
(6)

Square of equation (3) gives

(a -b )^2 = 4

Expand LHS

a^2 - 2ab +b^2 = 4
(7)

Substract (7) from (6) and solve for ab

ab = \dfrac{1}{3}(\dfrac{2\cdot 109}{2} - 2^2 ) = 35
(8)

Substituting ab to (6) gives

a^2+b^2 = 109 - ab = 74
(9)

Construct a perfect square binomial with (9) and (8)

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 74 +2\cdot 35 = 144

Taking square root of both sides gives

a +b = \pm12
(10)

Now we get two systems of equations

Case 1

\begin{cases} a+b = 12 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= 7 \\ b =11 \end{cases}

Then

x = a^3 - 109 = 7^3 - 109 = 234

Case 2

\begin{cases} a+b = -12 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= -5 \\ b =-7 \end{cases}

Then

x = a^3 - 109 = -5^3 - 109 = -234

In summary, we get two solutions for x.

x = 234 \text{ or }x = -234

Steven Zheng posted 4 months ago

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