Solving a radical equation

vaironxxrd

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Oct 30, 2011
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Hello forum, Vaironxxrd here.
I'm having some problems with my Algebra II Homework.

Problem: 27x+41=7\displaystyle 2\sqrt{7x + 4} -1 = 7

What I believe is the solution

(27x+4)2=(8)2\displaystyle (2\sqrt{7x + 4})^2 = (8)^2

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

4+7x=60\displaystyle 4 + 7x = 60

7x=56\displaystyle 7x = 56 =

x=8\displaystyle x = 8
 
Plug your solution back in and see if it equals 7. In doing so, it does not equal 7

You have a small mistake.

(27x+4)2=4(7x+4)\displaystyle (2\sqrt{7x+4})^{2}=4(7x+4), not 4+7x+4\displaystyle 4+7x+4
 
Plug your solution back in and see if it equals 7. In doing so, it does not equal 7

You have a small mistake.

(27x+4)2=4(7x+4)\displaystyle (2\sqrt{7x+4})^{2}=4(7x+4), not 4+7x+4\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...

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

28x + 16 = 64

28x=48\displaystyle 28x=48

x=4828\displaystyle x = \frac{48}{28}

x=127\displaystyle x = \frac{12}{7}
 
Last edited:
Thanks for telling me that, I did not know that rule.

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

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

28x + 16 = 64

28x=48\displaystyle 28x=48

x=4828\displaystyle x = \frac{48}{28}

x=127\displaystyle x = \frac{12}{7}


That is not wrong. x=12/7 is correct. Plug it back in and you see.
 
Hello forum, Vaironxxrd here.
I'm having some problems with my Algebra II Homework.

Problem: 27x+41=7\displaystyle 2\sqrt{7x + 4} -1 = 7

What I believe is the solution

(27x+4)2=(8)2\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:


27x+42 = 82\displaystyle \dfrac{2\sqrt{7x + 4}}{2} \ = \ \dfrac{8}{2}


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


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


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


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