# Finding value of a constant c in standard form of quadratic equation

• Nov 28th 2011, 02:47 PM
benny92000
Finding value of a constant c in standard form of quadratic equation
The equation is 16x^(2) -24x+c=0 What is c, given that one root is twice the other root?

I made one root r and the other root 2r. I know the -b/a and the c/a relationships (-b/a is the sum of the roots and c/a is the product). Thanks for the help.
• Nov 28th 2011, 02:53 PM
Plato
Re: Finding value of a constant c in standard form of quadratic equation
Quote:

Originally Posted by benny92000
The equation is 16x^(2) -24x+c=0 What is c, given that one root is twice the other root?

The sum of the roots is $\displaystyle \frac{24}{16}~.$

The product of the roots is $\displaystyle \frac{c}{16}~.$
• Nov 28th 2011, 02:56 PM
TKHunny
Re: Finding value of a constant c in standard form of quadratic equation
Let's just think about this for a moment. If there are two roots, a and b, what does it look like? Where did those relationships come from?

(x-a)*(x-b) = x^2 - (a+b)x + ab

This should suggest that the produce to the roots IS the constant term! Use this to your advantage.

16x^2 - 24x + c = 16(x^2 - (3/2)x + (c/16))

Using your notation, r * 2r = c/16 and you are almost done.

Also, notice how the opposite of the coefficient on the linear term is the SUM of the roots!

Using your notation, r + 2r = 3/2.

Go ahead and build structures and explore the problem. There is just no need to KNOW how to proceed before you think you are solving the problem. Invent a solution!
• Nov 28th 2011, 02:57 PM
benny92000
Re: Finding value of a constant c in standard form of quadratic equation
And how would we solve for the roots themselves?

Edit: I posted this at the same time as you. Give me a second to read :)

I came up with C=8 and the roots are 1/2 and 1.Touche..