SOLUTION: 7) The correct solution to 3=2√x+x is:
a. x=1 and x=3
b. x=1 and x= 4
c. x=1 and x=9
I can not figure this out...this is what I keep getting:
3=2√x+x
9=4x+
Algebra ->
Radicals
-> SOLUTION: 7) The correct solution to 3=2√x+x is:
a. x=1 and x=3
b. x=1 and x= 4
c. x=1 and x=9
I can not figure this out...this is what I keep getting:
3=2√x+x
9=4x+
Log On
a. x=1 and x=3
b. x=1 and x= 4
c. x=1 and x=9
I can not figure this out...this is what I keep getting:
3=2√x+x
9=4x+x²
x²+4x-9=0
And my answer does not factor into any of the possible answers, so I am completely stuck. Can you help me? Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website! you have done THE classic error when squaring. You cannot square each term...you square EACH SIDE, so you would have:
which will be horrible to expand.
The secret is to get the term with the square root in it by itself and THEN square both sides, as follows.
Now square both sides, to give
(x-9)(x-1) = 0
--> x-9=0 OR x-1=0
so x=9 or x=1
