# Quadratic equations: what possible values of k.

• Jul 17th 2011, 06:29 AM
rcs
Quadratic equations: what possible values of k.
How is it possible to solve this ? Determine values(s) of k for which
1. kx^2 – x – 4 = 0
2. 3x^2 + 5x + k = 0
3. 4x^2 – 3x + k = 0
4. kx^2 + 3x - 7 = 0
5. kx^2 - x - 3 = 0

Thank you
• Jul 17th 2011, 06:38 AM
Sudharaka
Re: what possible values of k.
Quote:

Originally Posted by rcs
How is it possible to solve this ? Determine values(s) of k for which
1. kx^2 – x – 4 = 0
2. 3x^2 + 5x + k = 0
3. 4x^2 – 3x + k = 0
4. kx^2 + 3x - 7 = 0
5. kx^2 - x - 3 = 0

Thank you

Hi rcs,

This system is inconsistent. Since,

$kx^{2}-x-4=0\mbox{ and }kx^{2}-x-3=0\Rightarrow 4=3$ ; which is obviously a contradiction. Hence the system does not have any solution.
• Jul 17th 2011, 07:28 AM
Soroban
Re: what possible values of k.
Hello, rcs!

Quote:

$\text{Determine values(s) of }k\text{ for which:}$

. . $\begin{array}{ccccc}1.& kx^2 – x – 4 &=& 0 \\ 2.& 3x^2 + 5x + k &=& 0 \\ 3. & 4x^2 – 3x + k &=& 0 \\ 4.& kx^2 + 3x - 7 &=& 0 \\ 5. & kx^2 - x - 3 &=& 0 \end{array}$
.

The problem is not clearly stated.

If complex roots are permitted, any value of $k$ can be used.

If the roots must be real, the discriminant must be nonnegative.

If the roots must be rational, the discriminant must be a square.