Question

What is the solution to the equation sqrt(x^2+2x-25)=sqrt(x+5)?

Answer 1

We have to decide if the square root is the radical sign, meaning the principal square root, or a multivalued square root.

Let's assume this is the radical sign applied to the radicands, indicating both sides are the principal square root.

$\sqrt{x^2 + 2x - 25}= \sqrt{x+5}$

In this case, dealing with the principal values, it's ok to square both sides. Both sides are positive (or a positive number times i); we won't introduce extraneous roots.

$x^2 + 2x - 25 = x+5$

$x^2+ x -30 = 0$

$(x-5)(x+6)=0$

Answer: x=5 or x=-6

Check:

$\sqrt{5^2+2(5)-25}=\sqrt{10}$

$\sqrt{5+5}=\sqrt{10} \quad\checkmark$

$\sqrt{(-6)^2 + 2(-6) - 25} = \sqrt{-1} = i$

$\sqrt{-6+5}=\sqrt{-1}=i \quad\checkmark$

I suppose it depends what grade you're in as to whether you'll accept x=6 as a solution.

Answer (restricting ourselves to real square roots): x=5

Answer 2

Answer: x=5 on edge

Step-by-step explanation:

I tried x= -6, x= 5 and it was incorrect.