# Thread: value of k and roots

1. ## value of k and roots

Can anybody help... how is it possible to find the root of 16x^2-24x+k = 0 ? when one root is 1/2 and find k?

it is possible to find the value of k if the sum and product of the roots are equal?

if im going to solve i understand that in the k is the c part of quadratic equation.
so,
k = c/a = k/16 then in this part i am stuck.
it made me solve for hours but still can't get the answer correctly.

thank you so much and more power.

2. ## Re: value of k and roots

Since 1/2 is a root then you know that $(2x -1)(Ax+B) = 16x^2-24x+k$ where A and B are constants.

Comparing coefficients

$x^2 \rightarrow 2A = 16 \text{ and } x^0 \rightarrow -B = k$

To find the value of B you can compare the coefficients of x. Once you have B it should be easy to find k

3. ## Re: value of k and roots

-B = k
and that is -24 = k is this correct sir? or should i have to make it -24 = r1 + r2; - 24 = 1/2 + r2 then -24 -1/2 = r2, then r2 = -24 1/2 which the other root now. Oh my gosh please hope this is right.

4. ## Re: value of k and roots

Comparing the coefficients of x gives $2B-A = -24$ and since we know A we have $2B-8 = -24$ which makes $B = -8$. Therefore I get $k=8$.

If we sub this back we get $16x^2-24x+8 = 8(2x-1)(x-1) = 0$

5. ## Re: value of k and roots

thank you sir... hope i can learn this more... i am not so good in mathematics

more power

6. ## Re: value of k and roots

Hello RCS,
given one root substitute it into your equation and find k =8
factor the new equation after simplification and find the other root ( 1)

bjh