How do you solve $\sqrt{x + 1} = \sqrt{x - 5}$ $sqrt(x+1) =sqrt(x-5)$ ?

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Answer 1 · 2018-03-23T16:33:24

No Solution

$\sqrt{x + 1} = \sqrt{x - 5}$ $sqrt(x+1) = sqrt(x-5)$

In order to eliminate a square root we can square the root.
Whatever you do on the left, you also do on the right.

${\sqrt{x + 1}}^{2} = {(\sqrt{x - 5})}^{2}$ $sqrt(x+1)^2 = (sqrt(x-5))^2$

This leaves us with the equation

$x + 1 = x - 5$ $x+1 = x-5$

Combine like terms by using the additive/subtractive inverse.

$cancelxcancel(+1) cancel(-x) cancel(-1)= cancelx -5 cancel(-x) -1$

This leaves

$0 != -6$
No Solution

How do you solve sqrt(x+1) =sqrt(x-5)√x+1=√x−5sqrt(x+1) =sqrt(x-5)?