# Thread: Solve for k: (Two different types of problems)

1. ## Solve for k: (Two different types of problems)

Hi all, thanks for taking time out of your day to have a look at my problems. I'm studying and need some help solving these two problems.

I need to solve for k:

Here are the problems:

1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4

2) kx^2 - 3x + 4 = 0 one root is twice the other.

I'm not really sure how to do these problems and am really greatful for your help! Btw, the "√" sign in the first problem is a square root sign (just in case someone got confused).

Thanks!

2. Originally Posted by OurIslamic

1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4

Thanks!
The other root is the complex conjugate. Take both linear roots and expand them back out. This should give you $\displaystyle k$.

Originally Posted by OurIslamic

2) kx^2 - 3x + 4 = 0 one root is twice the other.
Solve this using the quadratic formula, $\displaystyle x = \frac{3\pm\sqrt{9-16k}}{2k}$

Now ask what values for $\displaystyle k$ will make one root twice the other?

3. Originally Posted by OurIslamic
2) kx^2 - 3x + 4 = 0 one root is twice the other.
x = [3 + SQRT(9 - 16k)] / (2k) or [3 - SQRT(9 - 16k)] / (2k)

one root = twice the other:
3 + SQRT(9 - 16k) = 2[3 - SQRT(9 - 16k)]
3 + SQRT(9 - 16k) = 6 - 2SQRT(9 - 16k)
3SQRT(9 - 16k)] = 3
SQRT(9 - 16k)] = 1
9 - 16k = 1
16k = 8
k = 1/2

4. [QUOTE=OurIslamic;468155]Hi all, thanks for taking time out of your day to have a look at my problems. I'm studying and need some help solving these two problems.

I need to solve for k:

Here are the problems:

1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4
$\displaystyle 6(x- (1- i\sqrt{23})/4)(x- (1+ i\sqrt{23)/4$
$\displaystyle = 6((x-1/4)+ i\sqrt{23}/4)((x- 1/4)- i\sqrt{23}/4)$
is easier to calculate.

2) kx^2 - 3x + 4 = 0 one root is twice the other.
[/tex]kx^2- 3x+ 4= k(x- a)(x- 2a) [/tex]
Multiply that out and compare "corresponding coefficients to find both a and k.

I'm not really sure how to do these problems and am really greatful for your help! Btw, the "√" sign in the first problem is a square root sign (just in case someone got confused).

Thanks!