Thread: Solving For Variables In Terms Of Three Other Variables..?

Hello everyone!

I am new here, but I already have a quick question for you guys.

I am totally blanking out on how to do the following correctly:

This is the equation:

x=(-b+√(b^2-4ac))/2a

It's basically the quadratic formula, but with a plus instead of a plus/minus after the initial negative b term.

I am trying to solve for a,b, and c in terms of the other unused variables.

For example, I want to solve for a in terms of b, c, and x; b in terms of a, c, and x; and c in terms of a, b, and x.

I am not trying to find a numerical value for each of the three equations, though, just want to isolate each variable by itself on one side with all others on the other.

I have been working on this for hours and I am just not getting what I believe I should be getting. I feel like this should be cake but I am slowly driving myself mad trying to get this to work out.

If any of you guys can help me solve the three equations, your assistance would be greatly appreciated!

Ryan

Originally Posted by theryan

Hello everyone!

I am new here, but I already have a quick question for you guys.

I am totally blanking out on how to do the following correctly:

This is the equation:

x=(-b+√(b^2-4ac))/2a

It's basically the quadratic formula, but with a plus instead of a plus/minus after the initial negative b term.

I am trying to solve for a,b, and c in terms of the other unused variables.

For example, I want to solve for a in terms of b, c, and x; b in terms of a, c, and x; and c in terms of a, b, and x.

I am not trying to find a numerical value for each of the three equations, though, just want to isolate each variable by itself on one side with all others on the other.

I have been working on this for hours and I am just not getting what I believe I should be getting. I feel like this should be cake but I am slowly driving myself mad trying to get this to work out.

If any of you guys can help me solve the three equations, your assistance would be greatly appreciated!

Ryan

Well you know that this is the result when . This is the easier equation to solve for .

Originally Posted by Prove It

Well you know that this is the result when . This is the easier equation to solve for .

Okay, so I could just solve for a, b, and c with the quadratic equation you just supplied?

I got:

a = (-bx - c)/x^2

b = -ax - c

c = -ax^2 - bx

Would you agree with these?

Thank you very much!

Ryan

Originally Posted by theryan

Okay, so I could just solve for a, b, and c with the quadratic equation you just supplied?

I got:

a = (-bx - c)/x^2

b = -ax - c

c = -ax^2 - bx

Would you agree with these?

Thank you very much!

Ryan

I agree with your expressions for a and c. You need to double check your expression for b.

Originally Posted by theryan

This is the equation:
x=(-b+√(b^2-4ac))/2a

Good use of brackets, Ryan, but you missed a set:
x=(-b+√(b^2-4ac))/(2a)
See why?

20/2*4 = 40
20/(2*4) = 2.5