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Question

Difficulty: MediumRadical and Rational Equations

If xx satisfies the equation below, what is the value of x+5x + 5?

x−4=4x+5x - 4 = \sqrt{4x + 5}

Answer: 16

Answer

The correct answer is 16.
Squaring both sides of the equation x−4=4x+5x - 4 = \sqrt{4x + 5} results in (x−4)2=4x+5(x - 4)^2 = 4x + 5, which expands to x2−8x+16=4x+5x^2 - 8x + 16 = 4x + 5. Subtracting 4x+54x + 5 from both sides gives the quadratic equation x2−12x+11=0x^2 - 12x + 11 = 0. Factoring this quadratic equation yields (x−11)(x−1)=0(x - 11)(x - 1) = 0, giving candidate solutions of x=11x = 11 and x=1x = 1. Checking these solutions in the original equation shows that x=11x = 11 is valid (7=77 = 7), whereas x=1x = 1 is extraneous (−3=3-3 = 3 is false). Therefore, the only real solution is x=11x = 11, and the value of the expression x+5x + 5 is 11+5=1611 + 5 = 16.

Step-by-Step Solution

1
Square both sides of the equation to remove the radical.
(x−4)2=4x+5(x - 4)^2 = 4x + 5
Squaring a square root isolates the expression under the radical.
2
Expand the squared binomial on the left side.
x2−8x+16=4x+5x^2 - 8x + 16 = 4x + 5
Applying the distributive property to (x−4)(x−4)(x - 4)(x - 4) yields x2−8x+16x^2 - 8x + 16.
3
Rearrange the equation to set it equal to zero.
x2−12x+11=0x^2 - 12x + 11 = 0
Subtracting 4x+54x + 5 from both sides simplifies the equation into standard quadratic form.
4
Factor the quadratic equation.
(x−11)(x−1)=0(x - 11)(x - 1) = 0
Finding two numbers that multiply to 11 and add to -12 gives -11 and -1.
5
Solve for the candidate values of x.
x=11x = 11 and x=1x = 1
Setting each factor equal to zero yields the possible solutions.
6
Check the candidate values in the original equation to identify any extraneous solutions.
x=11x = 11 is valid; x=1x = 1 is extraneous.
Substituting x=1x = 1 results in −3=3-3 = 3, which is false because the principal square root is always non-negative.
7
Evaluate the target expression using the valid solution.
11+5=1611 + 5 = 16
The question asks for the value of x+5x + 5, so we substitute the only valid solution, x=11x = 11.

Key Concept

Solving radical equations by squaring both sides and checking for extraneous solutions.
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Topics

Radical and Rational Equations

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