Solve x^.5 = 6 - x

Nov 2009
76
1
What I have done is squared both sides which gives
x=(6-x)^2 =
0 = x^2-13x+36
0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

Is there a better way of doing this??
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
What I have done is squared both sides which gives
x=(6-x)^2 =
0 = x^2-13x+36
0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

Is there a better way of doing this??
\(\displaystyle \sqrt{x} = 6 - x\)

\(\displaystyle x = (6 - x)^2\)

\(\displaystyle x = x^2 - 12x + 36\)

\(\displaystyle 0 = x^2 - 13x + 36\)

\(\displaystyle 0 = x^2 - 9x - 4x + 36\)

\(\displaystyle 0 = x(x - 9) - 4(x - 9)\)

\(\displaystyle 0 = (x - 9)(x - 4)\)


So \(\displaystyle x - 9 = 0\) or \(\displaystyle x - 4 = 0\).

Therefore \(\displaystyle x = 9\) or \(\displaystyle x = 4\).


By the way, both values of \(\displaystyle x\) fit the function...
 
Nov 2009
76
1
I see your point. Thanks for that.
 
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