Resolving Algebra Problem: Finding Real Solutions for \sqrt{2x-3} + x = 3

[sauri]

I got this algebra problem that I have been trying to get around but the answer I get is always different from the book.

It says: Find all the real solutions, if any of each equation.

21). $$\sqrt{2x-3} +x=3$$.

I tried to resolve it using the following method:

2x-3+x^2=9
(2x-12)+(x^2-9)=0
2(x-6)+(x+3)(x-3)=0
(x+3) and (x-3) will cancel each other so;
2x-12=0
x=12/2
x=6.

However, this is wrong and I know it, cause the answer does not fit the equation. So can anyone help me point what's wrong with what I did?

 

[HallsofIvy]

Yes, that's wrong! (a+ b)² is not a²+ b² so $(\sqrt{2x-3}+x)^2$ is not 2x-3+ x!

I recommend you subtract x from both sides of th equation to get
$\sqrt{2x-3}= 3- x$, with only the square root on one side and then square both sides.

 

[sauri]

If I use $\sqrt{2x-3}= 3- x$, I then worked it out to
2x-3=9-6x+x^2
2x+6x-x^2=9+3
8x-x^2-12=0
So I re-wrote as;
-x^2+8x-12=0
(x+6)=0 or (x+2)=0

but then the answers add up as x=-6 or x=-2. and that can't be right..

 

[daveb]

You didn't quite factor it correctly.

 

[Nimz]

Try factoring -x^2+8x-12=0 again. You may want to move everything to the right hand side to make it a bit easier.

 

[sauri]

O.k, this time I tried the copleting the square method:

-x^2+8x-12=0
-x^2+8x=12
(x+4)^2=12+16
$(x+4)=\sqrt28$
$4+\sqrt28$ or $4-\sqrt28$

 

[sauri]

The answers vary between -1.29 and 9.29, Still off

 

[daveb]

-x^2 does not factor as (x+...)(x+...). That's your mistake.

 

[HallsofIvy]

Rewrite -x²+ 8x- 12= 0 as x²- 8x+ 12= 0 by multiplying both sides by -1 (or "move everything to the right hand side" as was suggested before.

x²- 8x+ 12= (x- ?)(x- ?).

Of course, be sure to check solutions to the polynomial equation in the original equation!

 

[sauri]

ah, I got it. You end up with (x- 6)(x- 2).
and only x=2 would apply for the equation.
Thanks