Question

what are the solutions to the equation \\(\sqrt{x^2 + 2x - 10} = \sqrt{x + 20}\\)? \\(x = 5, x = 6\\) \\(x = -6, x = -5\\) \\(x = -6, x = 5\\) \\(x = -5, x = 6\\)

Explanation:

Step1: Square both sides

To eliminate the square roots, square both sides of the equation \(\sqrt{x^2 + 2x - 10}=\sqrt{x + 20}\). This gives \(x^2 + 2x - 10=x + 20\).

Step2: Rearrange into quadratic

Subtract \(x + 20\) from both sides to get a quadratic equation: \(x^2+2x - 10-(x + 20)=0\), which simplifies to \(x^2+x - 30 = 0\).

Step3: Factor the quadratic

Factor \(x^2+x - 30\). We need two numbers that multiply to \(-30\) and add to \(1\). Those numbers are \(6\) and \(-5\), so the equation factors as \((x + 6)(x - 5)=0\).

Step4: Solve for x

Set each factor equal to zero: \(x+6 = 0\) gives \(x=-6\), and \(x - 5=0\) gives \(x = 5\).

Step5: Check solutions

• For \(x=-6\): Check the original equation. Left side: \(\sqrt{(-6)^2+2(-6)-10}=\sqrt{36 - 12 - 10}=\sqrt{14}\). Right side: \(\sqrt{-6 + 20}=\sqrt{14}\). So \(x=-6\) is valid.
• For \(x = 5\): Left side: \(\sqrt{5^2+2(5)-10}=\sqrt{25 + 10 - 10}=\sqrt{25}=5\). Right side: \(\sqrt{5 + 20}=\sqrt{25}=5\). So \(x = 5\) is valid.

Answer:

\(x=-6,x = 5\) (corresponding to the third option: \(x=-6,x = 5\))