# square roots

• Apr 20th 2008, 01:28 PM
square roots
1.the square root of y+4 -2=3

2.the cubed root of 6x+9 +5=2
the space implies that the other numbers are not under the square root
• Apr 20th 2008, 01:53 PM
any help would be greatly appreciated
• Apr 20th 2008, 02:33 PM
bobak
Quote:

1. $\displaystyle \sqrt{y+4} - 2 = 3$

what is wrong with $\displaystyle \sqrt{y+4} = 5$
$\displaystyle y+4 = 25$
$\displaystyle y = 21$

Or am i reading you question wrong ?

Bobak
• Apr 20th 2008, 02:44 PM
$\displaystyle \sqrt{y+4}-2 = 3$
• Apr 20th 2008, 02:53 PM
this is how it was suposed to read
• Apr 20th 2008, 11:07 PM
earboth
Quote:

this is how it was suposed to read

Your original equation is: ....... $\displaystyle \sqrt{y+4}-3=2$ ....... Add 3 on both sides.

You'll get bobak's equation.

But since squaring both sides of an equation is not an equivalent operation you must prove if the solution is valid or not. Plug in the solution into the original equation and prove if the equation is true:

$\displaystyle \sqrt{21+4}-3 \ \buildrel {\rm?} \over {\rm=} \ 2$

$\displaystyle 5-3 \buildrel {\rm?} \over {\rm=} \ 2$

$\displaystyle 2 \ \buildrel {\rm!} \over {\rm=} \ 2$....... So y = 21 is a valid solution.

to #2:

$\displaystyle \sqrt[3]{6x+9} + 5 = 2~\iff~ \sqrt[3]{6x+9}=-3$

Cube both sides. You'll get:

$\displaystyle 6x+9 = -27 ~\implies~ 6x =-36~\implies~x = -6$

Plug in this value into the original equation:

$\displaystyle \sqrt[3]{6 \cdot (-6)+9}+5 \buildrel {\rm?} \over {\rm=} 2~\iff~\sqrt[3]{-27}+5 \buildrel {\rm?} \over {\rm=} 2~\iff~ -3+5 \buildrel {\rm?} \over {\rm=} 2$....... which is obviously true.

And therefore x = -6 is a solution of this equation.