Question

what is the solution to the equation \\(\sqrt{x^2 + 2x - 25} = \sqrt{x + 5}\\)? assume the range is all real numbers
\\(\bigcirc\\ x = -6\\)
\\(\bigcirc\\ x = 5\\)
\\(\bigcirc\\ x = 6, x = -5\\)
\\(\bigcirc\\ x = -6, x = 5\\)

Explanation:

Step1: Square both sides

Since both sides are square roots, we can square both sides to eliminate the square roots. So we get \(x^{2}+2x - 25=x + 5\).

Step2: Rearrange into quadratic equation

Subtract \(x + 5\) from both sides to get a quadratic equation: \(x^{2}+2x-25-(x + 5)=0\), which simplifies to \(x^{2}+x-30 = 0\).

Step3: Factor the quadratic equation

Factor the quadratic equation: \(x^{2}+x - 30=(x + 6)(x - 5)=0\).

Step4: Solve for x

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

Step5: Check for extraneous solutions

• For \(x=-6\): Check the original equation. Left side: \(\sqrt{(-6)^{2}+2\times(-6)-25}=\sqrt{36-12 - 25}=\sqrt{-1}\), which is not a real number. So \(x=-6\) is extraneous.
• For \(x = 5\): Left side: \(\sqrt{5^{2}+2\times5-25}=\sqrt{25 + 10-25}=\sqrt{10}\); Right side: \(\sqrt{5 + 5}=\sqrt{10}\). So \(x = 5\) is a valid solution. Wait, but wait, maybe I made a mistake in step 5. Wait, let's re - check \(x=-6\): \((-6)^{2}+2\times(-6)-25=36-12 - 25=-1\), which is negative, so the square root is not real. For \(x = 5\): \(25 + 10-25 = 10\), positive. Wait, but the original options have \(x=-6,x = 5\) and \(x = 5\) etc. Wait, maybe I made a mistake in step 1. Wait, the original equation is \(\sqrt{x^{2}+2x-25}=\sqrt{x + 5}\). So the expressions inside the square roots must be non - negative.

For \(x=-6\): Inside the left square root: \(x^{2}+2x-25=36-12 - 25=-1\lt0\), so it's invalid. For \(x = 5\): Inside left square root: \(25 + 10-25 = 10\gt0\), inside right square root: \(5 + 5=10\gt0\). Now, let's re - do the quadratic solution. Wait, when we squared both sides, we might have introduced an extraneous solution. But let's check the options. The option with \(x = 5\) is there. Wait, but the option \(x=-6,x = 5\) is also there. But since \(x=-6\) gives a non - real number in the square root, it's extraneous. But wait, maybe I made a mistake in calculating \(x=-6\)'s inside the square root. Wait, \(x=-6\): \(x^{2}+2x-25=(-6)^2+2*(-6)-25=36 - 12-25=-1\), yes, negative. So only \(x = 5\) is valid? But the options: the second option is \(x = 5\). Wait, but let's check the quadratic equation again. \(x^{2}+x - 30=0\), solutions \(x=\frac{-1\pm\sqrt{1+120}}{2}=\frac{-1\pm11}{2}\). So \(x=\frac{-1 + 11}{2}=5\), \(x=\frac{-1-11}{2}=-6\). But we have to check the domain. The domain requires \(x^{2}+2x-25\geq0\) and \(x + 5\geq0\).
From \(x + 5\geq0\), we have \(x\geq - 5\).
For \(x=-6\), \(x\geq - 5\) is not satisfied. For \(x = 5\), \(x\geq - 5\) is satisfied. And \(x^{2}+2x-25=10\geq0\). So the only valid solution is \(x = 5\). But the option \(x = 5\) is present. Wait, but maybe the problem says "assume the range is all real numbers", maybe it means that we just solve the equation algebraically first and then check. Wait, but in the original options, the option \(x = 5\) is one of the options.

Answer:

B. \(x = 5\)