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Chapter 0: Problem 124

Solve each radical equation. Check all proposed solutions. $$\sqrt{2 x+15}-6=x$$

Short Answer

Expert verified
The solution of the equation \(\sqrt{2 x+15}-6=x\) is \(x=-3\).

Step by step solution

01

Isolate the root

The first step is to isolate the square root on one side of the equation by moving 6 to the right side of the equation. It becomes: \\[\sqrt{2x+15} = x+6\\]
02

Square both sides

After isolating the square root, square both sides of the equation to eliminate the square root. This results in: \\[2x+15 = (x+6)^2\\] Apply the formula for squaring a binomial to expand the right side: \\[2x+15 = x^2+12x+36\\]
03

Rearrange the equation to a quadratic equation format

Rearrange the equation into the standard form of a quadratic equation, i.e., \\[ ax^2+bx+c=0.\\] Thus, transfer 2x and 15 to the right side: \\[x^2+12x-2x+36-15=0\\] Simplifying, we get: \\[x^2+10x+21=0\\]
04

Solve the quadratic equation

Solving this equation by factoring, we have: \\[x^2+10x+21=(x+7)(x+3)=0\\] Thus, the possible solutions are: \\[x=-7, -3\\]
05

Check the proposed solutions

To ensure that these solutions are not extraneous, substitute them into the original equation and check if both sides are equal. Substituting \(x=-7\) into the original equation, \\[\sqrt{2(-7)+15}\neq -7+6.\\] Those two sides are not equal, therefore \(x=-7\) is an extraneous solution. Substituting \(x=-3\) into the original equation, \\[\sqrt{2(-3)+15}= -3+6.\\] The left and right sides are equal, hence \(x=-3\) is a valid solution.

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Most popular questions from this chapter

Will help you prepare for the material covered in the next section. If the width of a rectangle is represented by \(x\) and the length is represented by \(x+200\), write a simplified algebraic expression that models the rectangle's perimeter.

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