Question

Need help with solving radical equations

Answer 1

were is it?

Answer 2

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

Answer 3

This is an example question

Answer 4

I nee steps on how to solve

Answer 5

3

Answer 6

I know the answer -_- Need help with steps to solve the problem

Answer 7

lol

Answer 8

\[x = \sqrt{4 - x} + 2\]
1) subtract \(2\) to isolate the radical, get
\[x-2=\sqrt{4-x}\]

Answer 9

then get rid of the radical by squaring both sides (carefully)
\[(x-2)^2=4-x\\
x^2-4x+4=4-x\]

Answer 10

solve the quadratic equation by setting this equal to zero
\[x^2-3x=0\]

Answer 11

Thanks @satellite73 For this equation 1+sqrt{x-3}=sqrt{2x-6} what is the first step?

Answer 12

\[1+\sqrt{x-3}=\sqrt{2x-6}\]

Answer 13

this one is going to be a drag because you will have to square twice

Answer 14

should I isolate the 1 and bring all the radicals on one side first?

Answer 15

no that will make matters worse
\[(1+\sqrt{x-3})^2=\sqrt{2x-6}^2\]

Answer 16

square both sides

Answer 17

just be careful when you square
you should bet
\[1+2\sqrt{x-3}+x-3=2x-6\]

Answer 18

then isolate the radical and square again

Answer 19

Please explain how you squared the left side

Answer 20

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

Answer 21

that is precisely why i wrote "carefully" as
\[(a+b)^2=a^2+2ab+b^2\]

Answer 22

after that step do I square both sides again?

Answer 23

we are here
\[1+2\sqrt{x-3}+x-3=2x-6\] right? now we need to isolate the radical
don't square now, it will be a huge mess

Answer 24

\[1+2\sqrt{x-3}+x-3=2x-6\\
2\sqrt{x-3}+x-2=2x-6\\
2\sqrt{x-3}=x+4\]

Answer 25

NOW square again

Answer 26

I believe right is supposed to be x-4

Answer 27

yes you are correct

Answer 28

\[\left( 2\sqrt{x-3} \right)^{2} = (x-4)^{2}\]
\[4x + 12 = x ^{2} - 8x +16\]

Answer 29

approximately 3 is the answer