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:


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


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


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


\(\displaystyle 7x + 4 = 16\)


\(\displaystyle \text{And continue . . . }\)