what is the solution to the equation $sqrt{x^{2}+2x - 25}=sqrt{x + 5}$? assume the range is all real…

Explanation:

Step1: Square both sides

Since $\sqrt{a}=\sqrt{b}$ implies $a = b$ (for non - negative $a$ and $b$), we have $x^{2}+2x - 25=x + 5$.

Step2: Rearrange to quadratic form

Move all terms to one side: $x^{2}+2x-x-25 - 5=0$, which simplifies to $x^{2}+x-30 = 0$.

Step3: Factor the quadratic

We factor $x^{2}+x - 30$ as $(x + 6)(x - 5)=0$.

Step4: Solve for x

Setting each factor equal to zero gives $x+6=0$ or $x - 5=0$, so $x=-6$ or $x = 5$.

Step5: Check for extraneous solutions

For $x=-6$: Left - hand side: $\sqrt{(-6)^{2}+2\times(-6)-25}=\sqrt{36-12 - 25}=\sqrt{-1}$, which is not a real number in the set of real numbers. For $x = 5$: Left - hand side: $\sqrt{5^{2}+2\times5-25}=\sqrt{25 + 10-25}=\sqrt{10}$, right - hand side: $\sqrt{5 + 5}=\sqrt{10}$.

Answer:

$x = 5$