• Jan 11th 2007, 02:28 PM
bobbyfisher
radical y squared + 2 = y-1

• Jan 11th 2007, 05:26 PM
topsquark
Quote:

Originally Posted by bobbyfisher
radical y squared + 2 = y-1

I'm not quite sure of what the equation is. It just doesn't make sense to me. What you have written is either
$\displaystyle \sqrt{y^2} + 2 = y - 1$

or

$\displaystyle (\sqrt{y})^2 + 2 = y - 1$

Neither of which should pose a problem.

I'm going to assume that you wrote it wrong and meant:
$\displaystyle \sqrt{y} + 2 = y - 1$

$\displaystyle \sqrt{y} = y - 3$ <-- Square both sides

$\displaystyle y = (y - 3)^2$

$\displaystyle y = y^2 - 6y + 9$

$\displaystyle y^2 - 7y + 9 = 0$

which you can solve using the quadratic formula:
$\displaystyle y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(9)}}{2 \cdot 1}$

$\displaystyle y = \frac{7 \pm \sqrt{13}}{2}$

The bad news is that whenever you square both sides of an equation you should always check to make sure you haven't added a new solution and these forms are messy to substitute back into the original equations. I'll give you peace of mind by telling you that only the "+" solution is correct.

-Dan
• Jan 11th 2007, 09:03 PM
CaptainBlack
Quote:

Originally Posted by bobbyfisher
radical y squared + 2 = y-1

If:

$\displaystyle \sqrt{y^2+2}=y-1$,

then:

$\displaystyle y^2+2=(y-1)^2=y^2-2y+1$

which simplifies to:

$\displaystyle 2y+1=0$

or $\displaystyle y=-1/2$.

Now the squaring can introduce spurious solutions so this needs to be
substituted back into the original equation to check that it is a solution.

Doing this we find it is not a solution so the equation has no solutions!

RonL