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

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- Apr 20th 2008, 01:28 PMimbadatmathsquare 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 PMimbadatmath
any help would be greatly appreciated

- Apr 20th 2008, 02:33 PMbobak
- Apr 20th 2008, 02:44 PMimbadatmath
$\displaystyle

\sqrt{y+4}-2 = 3

$ - Apr 20th 2008, 02:53 PMimbadatmath
this is how it was suposed to read

- Apr 20th 2008, 11:07 PMearboth
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.