Basic rule

Chizum · #1 $Old$ January 6th, 2009, 11:26 PM

I'm in calculus but there is a basic precalculus rule that I'm shady on and need to use.

If I had $y^2 = 3 - x$ and I wanted to solve for y, would I get y = +/- the square root of x - 3 ?

I forget if that is the proper situation when you have to use +/-

IF it is, then I need to set it equal to x - 1 and solve for x. How would that solve when the +/- is involved?

Mush · #2 $Old$ January 6th, 2009, 11:32 PM

Quote:

Originally Posted by Chizum $View Post$

I'm in calculus but there is a basic precalculus rule that I'm shady on and need to use.

If I had $y^2 = 3 - x$ and I wanted to solve for y, would I get y = +/- the square root of x - 3 ?

I forget if that is the proper situation when you have to use +/-

IF it is, then I need to set it equal to x - 1 and solve for x. How would that solve when the +/- is involved?

Yes. Whenever you have an equation of the form $y^2 = something$ , then y is ALWAYS given by $y = \pm \sqrt{something}$ .

This is because x^2 = (-x)^2, and hence if you were to do the inverse, you wouldn't know whether x or -x was the original term, so we include both as solutions!

This isn't just the rule, but in fact, it is the rule for any EVEN exponent.

So if you had $y^4, y^6, y^{2023424}, y^{99999999994}$ , then your answer would always be $\pm$ .

Chizum · #3 $Old$ January 6th, 2009, 11:42 PM

Thanks for the thorough answer.

How would I solve $x - 1 = +/- \sqrt{3 - x}$ for x?

The +/- throws me off.

DaveDammit · #4 $Old$ January 6th, 2009, 11:44 PM

$\sqrt{y^2} = \sqrt{3 - x}$

= $y = +/- \sqrt{3 - x}$

The equation already states that you're solving for $y$ . If you solved for $x$ why not just solve in the beginning?

$y^2 = 3 - x$

= $\sqrt{y^2 + x} = +/- \sqrt{3}$

Can you solve the rest for $x$ ?

If this is wrong maybe someone can correct me but this seems like the easiest way to me...

Chizum · #5 $Old$ January 6th, 2009, 11:53 PM

I had to solve two equations for y and then equate those values and solve and plug the result(s) back into the one of the original equations (to find the points of intersection of these two equations if graphed). The equations are $x = 3 - y^2$ and $y = x - 1$ . I'm on the step where I need to equate them and solve. I just don't remember at all going over how to solve equalities when +/- is involved.

Mush · #6 $Old$ January 7th, 2009, 12:01 AM

Quote:

Originally Posted by Chizum $View Post$

I had to solve two equations for y and then equate those values and solve and plug the result(s) back into the one of the original equations (to find the points of intersection of these two equations if graphed). The equations are $x = 3 - y^2$ and $y = x - 1$ . I'm on the step where I need to equate them and solve. I just don't remember at all going over how to solve equalities when +/- is involved.

$x= 3-y^2$ . Just plug this into the other equation $y = x-1$ , to get:

$y = 3-y^2-1$

Rearrange:

$y^2+y-2 = 0$

However for future reference, if you have the equation $y = \pm \sqrt{3-x}$ and the equation $y = x-1$ , then you would take the first equation and split it into two. And then do the usual process twice for each equation.

For example:

For $y = + \sqrt{3-x}$

$\sqrt{3-x}=x-1$

$3-x = (x-1)^2$

$3-x = x^2-2x+1$

$x^2-x-2 = 0$

For $y = - \sqrt{3-x}$

$-\sqrt{3-x}=x-1$

$\sqrt{3-x}=1-x$

$3-x = (1-x)^2$

$3-x = 1-2x+x^2$

$x^2-x-2 = 0$

It just so happens that in this case, both equations give identical results