Question

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

Answer 1

I tried it, but got something wrong, and I need someone to check it

Answer 2

subtract 3x from sides, then square both sides and note \((\sqrt{x})^2=x\)

Answer 3

9x^2+x-1=x^2+4
x^2+x=5
x^2+x-5=0

Answer 4

so like
x-1=x^2+4+9x^2

Answer 5

then to make it a quadratic don't I go
0=10x^2-x+5?

Answer 6

\(3x+\sqrt{x-1}=x+2\iff \sqrt{x-1}=-2x+2\iff (x-1)=(-2x+2)^2\)

Answer 7

<maybe that's where I messed up>
that would simplify to
x-1=4x+4

Answer 8

If you subtract 3x from both sides of your equation, you can then square both sides to get rid of the square root on the left side. \[3x+\sqrt{x-1}=x+2\]\[\sqrt{x-1} = x+2 - 3x\]\[(\sqrt{x-1})^2 = (-2x +2)^2\]

Answer 9

sorry *x^2
x-1=4x^2+4

Answer 10

0=4x^2-x+5
but since we can't factor it, ill do the -b+/-sqrt(b^2-4ac)/2a

Answer 11

Why are you even bothering to do that?

Answer 12

\(3x+\sqrt{x-1}=x+2\iff \sqrt{x-1}=-2x+2\iff (x-1)=(-2x+2)^2\)



\(x-1=4x^2-8x+4 \iff 4x^2-9x+5=0\)

Answer 13

this factors

Answer 14

I don't understand how you did that

Answer 15

He moved everything over to the right.

Answer 16

how did (-2x+2)^2 become 4x^2-8x+4?

Answer 17

won't that be
(-2)^2*(x)^2+(2)^2
4x^2+4

Answer 18

\[(a +b)^2 = a^2 +2ab+b^2\]

Answer 19

oh, I didn't see that sorry now that makes sense
then
4x^2-9x+5=0 is the quadratic answer, then
factoring it
(x-1)(4x-5)
1, 5/4 is the answer, just need to check it

Answer 20

mmhmm.

Answer 21

thank you!

Answer 22

\[4x^2-9x+5=0\]\[x^2 -9x +20=0\]\[(x-5)(x-4)=0\]\[\left(x-\frac{5}{4}\right)\left(x-\frac{4}{4}\right)=0\]\[(4x-5)(x-1)=0\]\(\checkmark\) good.