Question

Please help me solve this! I have done it over like 50 times!

Answer 1

\[1+\sqrt{3x+3}=\sqrt{7x+2}\] I am super confused, i keep doing it by squaring both sides and i get x=.5 but it should be x=2

Answer 2

Did you get this far?

( 1 + sqrt(3x+3))^2 = (sqrt(7x+2))^2

( 1 + sqrt(3x + 3))*( 1 + sqrt(3x + 3)) = 7x + 2

Answer 3

Oops, I think that is what I have been doing wrong. I thought i was just supposed to square them and it would look like this\[1+3x+3=7x+2\]

Answer 4

Original equation: 1 + sqrt(3x+3) = sqrt(7x+2)

square both sides
( 1 + sqrt(3x+3))^2 = (sqrt(7x+2))^2

( 1 + sqrt(3x + 3))*( 1 + sqrt(3x + 3)) = 7x + 2

1*1 + sqrt(3x+3)+ sqrt(3x+3) + (sqrt(3x+3))^2 = 7x + 2

1 + 2*sqrt(3x+3) + 3x+3 = 7x + 2

Answer 5

Hey @perl think you could tell me? I want to know just in case i get another problem like this.

Answer 6

I wanted to know if 1 was the reason this happened, it must have gotten deleted.

Answer 7

\[1+\sqrt{3x+3}=\sqrt{7x+2}\]\[(1+\sqrt{3x+3})^2 = (\sqrt{7x+2})^2\]\[\sf \text{recall that} ~(a+b)^2 = a^2 +2ab+b^2\]\[1+2\sqrt{3x+3}+3x+3=7x+2\]\[2\sqrt{3x+3}+3x+4=7x+2\]\[2\sqrt{3x+3}=4x-2\]\[\sqrt{3x+3}=2x-1\]\[(\sqrt{3x+3})^2=(2x-1)^2\]\[3x+3=4x^2-2(2x)(-1)+1\]\[3x+3=4x^2+4x+1\]

Answer 8

Original equation: 1 + sqrt(3x+3) = sqrt(7x+2)

square both sides
( 1 + sqrt(3x+3))^2 = (sqrt(7x+2))^2

( 1 + sqrt(3x + 3))*( 1 + sqrt(3x + 3)) = 7x + 2

1*1 + sqrt(3x+3)+ sqrt(3x+3) + (sqrt(3x+3))^2 = 7x + 2

1 + 2*sqrt(3x+3) + 3x+3 = 7x + 2


2*sqrt(3x+3) + 4 + 3x = 7x + 2


2*sqrt(3x+3) = 4x - 2

sqrt(3x+3) = 2x - 1

now square both sides

(sqrt(3x+3))^2 = (2x-1)^2

3x+3 = (2x-1)^2

3x + 3 = (2x-1)*(2x-1)

3x+3 = 4x^2 - 2x - 2x + 1

3x+3 = 4x^2 - 4x + 1

4x^2 - 4x - 3x + 1 - 3 = 0

4x^2 - 7x - 2 = 0

(4x + 1) ( x - 2) = 0

x = -1/4 , x = 2

plug both solutions into original equation

Answer 9

:)

Answer 10

Thanks guys, this helps a lot.

Answer 11

Let me check with you too:\[3x+3=4x^2-4x+1\]\[4x^2-7x-2=0\]\[x^2-7x-8=0\]\[(x-8)(x+1)=0\]\[\left(x-\frac{8}{4}\right)\left(x+\frac{1}{4}\right)=0\]\[(x-2)(4x+1)=0\]\[x-2 \implies x=2\]\[4x+1 \implies x=-\frac{1}{4}\]\[\checkmark\]