Solving a radical equation

[vaironxxrd]

Hello forum, Vaironxxrd here.
I'm having some problems with my Algebra II Homework.

Problem: $\displaystyle 2\sqrt{7x + 4} -1 = 7$

What I believe is the solution

$\displaystyle (2\sqrt{7x + 4})^2 = (8)^2$

$\displaystyle 4 + 7x +4 = 64$ =

$\displaystyle 4 + 7x = 60$

$\displaystyle 7x = 56$ =

$\displaystyle x = 8$

 

[galactus]

Plug your solution back in and see if it equals 7. In doing so, it does not equal 7

You have a small mistake.

$\displaystyle (2\sqrt{7x+4})^{2}=4(7x+4)$, not $\displaystyle 4+7x+4$

 

[vaironxxrd]

Plug your solution back in and see if it equals 7. In doing so, it does not equal 7

You have a small mistake.

$\displaystyle (2\sqrt{7x+4})^{2}=4(7x+4)$, not $\displaystyle 4+7x+4$

Thanks for telling me that, I did not know that rule.

When I solve it I still get a wrong result - Fix

I actually get the correct answer...

$\displaystyle 4(7x+4) = 64$

28x + 16 = 64

$\displaystyle 28x=48$

$\displaystyle x = \frac{48}{28}$

$\displaystyle x = \frac{12}{7}$

 

[galactus]

Thanks for telling me that, I did not know that rule.

When I solve it I still get a wrong result - Fix

$\displaystyle 4(7x+4) = 64$

28x + 16 = 64

$\displaystyle 28x=48$

$\displaystyle x = \frac{48}{28}$

$\displaystyle x = \frac{12}{7}$

That is not wrong. x=12/7 is correct. Plug it back in and you see.

 

[lookagain]

Hello forum, Vaironxxrd here.
I'm having some problems with my Algebra II Homework.

Problem: $\displaystyle 2\sqrt{7x + 4} -1 = 7$

What I believe is the solution

$\displaystyle (2\sqrt{7x + 4})^2 = (8)^2$

$\displaystyle \cdot$

$\displaystyle \cdot$

$\displaystyle \cdot$

vaironxxrd,

a recommendation is to divide each side by 2 before the squaring:


$\displaystyle \dfrac{2\sqrt{7x + 4}}{2} \ = \ \dfrac{8}{2}$


$\displaystyle \sqrt{7x + 4} \ = \ 4$


$\displaystyle (\sqrt{7x + 4})^2 \ = \ (4)^2$


$\displaystyle 7x + 4 = 16$


$\displaystyle \text{And continue . . . }$