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Question

Difficulty: MediumRadical and Rational Equations

If 2x+15−6=x\sqrt{2x + 15} - 6 = x, what is the value of x+5x + 5?

  1. A
    -3
  2. B
    -2
  3. 2Answer
  4. D
    8

Answer

The correct answer is 2.
The correct answer is 2. To solve the equation 2x+15−6=x\sqrt{2x + 15} - 6 = x, we first isolate the radical by adding 66 to both sides, obtaining 2x+15=x+6\sqrt{2x + 15} = x + 6. Squaring both sides yields 2x+15=x2+12x+362x + 15 = x^2 + 12x + 36. Rearranging the terms to form a quadratic equation gives x2+10x+21=0x^2 + 10x + 21 = 0, which factors as (x+7)(x+3)=0(x + 7)(x + 3) = 0. This gives two potential solutions: x=−7x = -7 and x=−3x = -3. We must check both potential solutions in the original equation. For x=−3x = -3, the equation holds true: 2(−3)+15−6=3−6=−3\sqrt{2(-3) + 15} - 6 = 3 - 6 = -3. For x=−7x = -7, it does not: 2(−7)+15−6=1−6=−5≠−7\sqrt{2(-7) + 15} - 6 = 1 - 6 = -5 \neq -7. Thus, x=−3x = -3 is the only valid solution. Finally, substituting this into the expression x+5x + 5 gives −3+5=2-3 + 5 = 2.

Step-by-Step Solution

1
Isolate the radical term on one side of the equation.
2x+15=x+6\sqrt{2x + 15} = x + 6
Before squaring both sides, the radical term must be isolated to avoid cross-terms on that side.
2
Square both sides of the equation to eliminate the radical.
2x+15=(x+6)2  ⟹  2x+15=x2+12x+362x + 15 = (x + 6)^2 \implies 2x + 15 = x^2 + 12x + 36
Squaring a square root cancels the radical, allowing us to solve the resulting algebraic equation.
3
Rearrange the equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
x2+10x+21=0x^2 + 10x + 21 = 0
Subtracting 2x2x and 1515 from both sides collects all terms on one side so the quadratic can be factored.
4
Factor the quadratic equation.
(x+7)(x+3)=0  ⟹  x=−7 or x=−3(x + 7)(x + 3) = 0 \implies x = -7 \text{ or } x = -3
Finding two numbers that multiply to 21 and add to 10 allows us to factor the quadratic and find potential solutions.
5
Substitute the potential solutions back into the original equation to check for extraneous solutions.
For x=−3x = -3, 2(−3)+15−6=−3\sqrt{2(-3) + 15} - 6 = -3 is true. For x=−7x = -7, 2(−7)+15−6=−7\sqrt{2(-7) + 15} - 6 = -7 is false (gives −5≠−7-5 \neq -7). Thus, x=−3x = -3 is the only valid solution.
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.
6
Evaluate the expression x+5x + 5 using the valid solution.
−3+5=2-3 + 5 = 2
The question asks for the value of x+5x + 5, not just the value of xx.

Key Concept

Solving radical equations by isolating the radical, squaring both sides, checking for extraneous solutions, and evaluating algebraic expressions.
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Topics

Radical and Rational Equations

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