Quadratic equations: what possible values of k.

**rcs** · July 17th 2011, 05:29 AM

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

**Sudharaka** · July 17th 2011, 05:38 AM

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.

**Soroban** · July 17th 2011, 06:28 AM

Hello, rcs!

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

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

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.

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