Question

Solve the equation:
2- 2x = sqrt{4 + x}

I tried solving this, and I got x = 9/4, 0.
But the answer to this is only x = 0. Why?

Answer 1

look at this plot

Answer 2

2-2(9/4) is a negative number. The Square root side can not equal a negative number. Therefore 9/4 is extraneous

Answer 3

Plug in 9/4 to verify and you'll see why that cant be an answer.....(the suspense!!!!)

Answer 4

mrjoselson ruined the ending <.< haha jk jk

Answer 5

:(

Answer 6

Why can't it equal a negative number? I thought square roots had that ± sign in front of them..

Answer 7

The function:
\[f(x) = \sqrt{x}\]
always returns the positive root. If it returned both the positive and negative, it wouldnt be a function anymore (it would fail the vertical line test).
Thats why when we want the negative root, or to show its existance we put a minus, or plus/minus in front.

Answer 8

i hope myininaya has a better explanation lol >.<

Answer 9

Here is a complete solution with explanations:

Answer 10

I wouldnt use the complete the square technique, thats a little too complicated for this problem:
\[4x^{2}-9x = x(4x-9)=0 \rightarrow x = 0, 4x-9=0 \rightarrow x = 0, x = \frac{9}{4}\]
Other than that, beautiful :) medals for everyone >.>

Answer 11

\[2-2x=\sqrt{4+x}\]
\[(2-2x)^2=(\sqrt{4+x})^2\]
\[(2-2x)^2=4+x\]
\[(2-2x)(2-2x)=4+x\]
\[4-4x-4x+4x^2=4+x\]
\[4x^2-8x+4=4+x\]
\[4x^2-8x-x+4-4=0\]
\[4x^2-9x=0\]
\[x(4x-9)=0\]
\[x=0, x=\frac{9}{4}\]
Check x=0:
\[2-2(0)=2=2=\sqrt{4+0}\]
so x=0 works!
Check \[x=\frac{9}{4}\]:
\[2-2(\frac{9}{4})=2-\frac{9}{2}=\frac{8}{4}-\frac{9}{4}=\frac{-1}{4}\]
\[\sqrt{4+\frac{9}{4}}=\sqrt{\frac{16}{4}+\frac{9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}\]
so we don't have the samething on both sides so \[x=\frac{9}{4}\]
is not a solution

Answer 12

whenever you raise both sides to an even power be sure to check your answers