## Ducky44 4 years ago √8x + 1 =5 show work

1. Thephillup

|dw:1320896121958:dw|

2. myininaya

is the problem unclear or is the problem exactly as stated?

3. myininaya

$\sqrt{8}x+1=5 => \sqrt{8}x=5-1=> \sqrt{8}x=4 => (\sqrt{8}x)^2=4^2$ $=>8x^2=16 => x^2=2 => x = \pm \sqrt{2}$ But only $x=\sqrt{2} \text{ is a solution}$

4. Ducky44

√x - 7 + 5 = 11 show work

5. myininaya

$\text{for some reason I think you meant } \sqrt{8x+1}=5$ square both sides $8x+1=5^2$ 8x+1=25 subtract 1 on both sides 8x=25-1 8x=24 divide by 8 on both sides x=3 check your result: $\sqrt{8(3)+1}=\sqrt{25}=5$ both sides are the same x=3 is a solution!

6. Thephillup

Wouldn't x=2 because 2*8=16 sqt16=4 4+1=5