Question

Can someone check my answer please?
rad(x+1)+4=2x
I got x=-5

Answer 1

\[\sqrt{x+1}+4=2x\]

Answer 2

\[2(\sqrt{x=1})=(2x-4)2\]
\[x+1=4x+16\]
\[-3x=15\]
\[x=-5\]

Answer 3

\(\bf (2x-4)^2\implies (2x-4)(2x-4) \ne 2^2x-4^2\)

Answer 4

Okay so my answer is wrong? @jdoe0001

Answer 5

yeap

Answer 6

Then what do I do?

Answer 7

is a quadratic...so it should have 2 values

Answer 8

How do I get the two values? I'm confused on where to start now

Answer 9

\(\bf \sqrt{x+1}+4=2x\implies x+1=(2x-4)^2
\\ \quad \\
x+1=(2x)^2-2(2x)(4)+(4)^2\implies x+1=4x^2-16x+16
\\ \quad \\
0=4x^2-17x+15\)

Answer 10

You messed up on your first/second step when you were trying to get rid of the radical sign

You multiplied both sides by 2
When you should have squared both sides.

Multiplying by 2 doesn't get rip do the radical sign

Answer 11

in order to get rid of the root, you raise the term by THAT MUCH
in this case, since the root is 2, you raise it by 2

Answer 12

okay thanks that makes more sense! @jdoe0001

Answer 13

yw