radicals

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January 11th 2007, 02:28 PM
bobbyfisher

radicals

radical y squared + 2 = y-1

i believe you have to square both sides, but i'm not sure. please help
January 11th 2007, 05:26 PM
topsquark

Quote:

Originally Posted by bobbyfisher

radical y squared + 2 = y-1

i believe you have to square both sides, but i'm not sure. please help

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

or

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

Neither of which should pose a problem.

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

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

$y = (y - 3)^2$

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

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

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

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

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
January 11th 2007, 09:03 PM
CaptainBlack

Quote:

Originally Posted by bobbyfisher

radical y squared + 2 = y-1

i believe you have to square both sides, but i'm not sure. please help

If:

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

then:

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

which simplifies to:

$2y+1=0$

or $y=-1/2$ .

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