Question

solve the equation by making an appropriate substitution? 2x-9 x(sq rt) -5=0

How do you do this?

Answer 1

\[2x - 9\sqrt{x} -5 = 0\]

Is it right?

Answer 2

Yes..

Answer 3

I take that back it is -2x not 2x, sorry..

Answer 4

My bad, you were right the first time. It' time to go to bed, I'm getting a brain-freeze... it is 2x

Answer 5

The substitution you want to make is:
\[y = \sqrt{x} \Rightarrow y^{2} = x\]
that way your equation becomes:
\[2x-9\sqrt{x}-5 =0\Rightarrow 2y^{2}-9y-5 = 0\]
Then you would use the quadratic formula

Answer 6

if someone else could finish, it would be greatly appreciated lol, im in the middle of eating and watching Sucker Punch >.>

Answer 7

2(-5)=-10(1)
-10+1=-9
we will replace -9y with -10y+1y
\[2y^2-10y+1y-5=0\]
\[2y(y-5)+1(y-5)=0\]
\[(y-5)(2y+1)=0\]
\[y=5, y=\frac{-1}{2}\]
but remember
\[y=\sqrt{x}\]
so we have
\[\sqrt{x}=5, \sqrt{x}=\frac{-1}{2}\]
to solve this we square both sides
\[x=25,x=\frac{1}{4}\]
we should check these since we raised both sides to an even power

Answer 8

so lets check x=25
\[2*25-9\sqrt{25}-5=50-9*5-5=50-45-5=5-5=0\]
since we have the samething on both sides then x=25 is a solution
and lets check x=1/4
\[2*\frac{1}{4}-9\sqrt{\frac{1}{4}}-5=\frac{1}{2}-9*\frac{1}{2}-5=\frac{-8}{2}-5=-4-5=-9\]
and x=1/4 is not a solution since both sides are not the same