square root{9x+67} = x + 5
I know the solution is 6, however i am stumped trying to figure out how to start. My apologies, i do not know exactly how to write square root signs, so whatever is in those funny looking parenthesis is what's inside the square root sign.

Question

square root{9x+67} = x + 5
I know the solution is 6, however i am stumped trying to figure out how to start. My apologies, i do not know exactly how to write square root signs, so whatever is in those funny looking parenthesis is what's inside the square root sign.

thesmartone · Answer

$\sf\Large\sqrt{9x+67} = x + 5$

You start off by squaring both sides

thesmartone · Answer

$\sf\Large\sqrt{9x+67} = x + 5$

$\sf\Large(\sqrt{9x+67})^2 = (x + 5)^2$

Simplifying that gets us:

$\sf\Large 9x+67 = (x + 5)^2$

So, expand (x+5)^2 what is it? :P

anonymous · Answer

x^2 +10x +25

thesmartone · Answer

correct

anonymous · Answer

thats where i got stuck

anonymous · Answer

because that x^2 doesnt go anywhere

thesmartone · Answer

$\sf\Large 9x+67 =x^2 + 10x + 25$

now make the equation set to 0

so bring 9x + 67 on the other side :)

subtract 9x and subtract 67 on both sides :)

anonymous · Answer

yeah, x^2 + x -42

anonymous · Answer

oh wait am i supposed to factor this?

jim_thompson5910 · Answer

yes or use the quadratic formula

thesmartone · Answer

^

anonymous · Answer

well factoring it gives me a weird number

anonymous · Answer

whats the quadratic formula?

jim_thompson5910 · Answer

to factor, find two numbers that

a) multiply to -42 (last term)
AND
b) add to 1 (middle coefficient)

anonymous · Answer

-6 7

jim_thompson5910 · Answer

yep

jim_thompson5910 · Answer

so it factors to (x-6)(x+7)

anonymous · Answer

oh i see, i typed it wrong on my calc

anonymous · Answer

so then how is it 6?

jim_thompson5910 · Answer

so if ` (x-6)(x+7) = 0`
then either `x-6 = 0` OR `x+7 = 0`

jim_thompson5910 · Answer

you then solve each equation for x

anonymous · Answer

if i set both to zero i will have solutions 6 and -7. but only one can be the answer i think

jim_thompson5910 · Answer

so the *possible* solutions are x = 6 or x = -7

what you need to do now is check each possible solution back into the original equation

jim_thompson5910 · Answer

Here's an example

say we had `sqrt(x) = -1`

squaring both sides gives `x = 1` but plugging that back into `sqrt(x) = -1` leads to `1 = -1` which is false. So `x = 1` is an extraneous solution (not a real solution at all)

anonymous · Answer

and -7 doesnt look like it works because it ends up being 2=-2

anonymous · Answer

so thats false and only 6 is true right?

jim_thompson5910 · Answer

yes, only x=6 is the true solution. x=-7 is extraneous

-7 satisfies every equation after you square both sides but it does NOT satisfy the original equation

anonymous · Answer

alright thanks! :)

jim_thompson5910 · Answer

no problem