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

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- Jul 17th 2011, 05:29 AMrcsQuadratic 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, 05:38 AMSudharakaRe: what possible values of k.
- Jul 17th 2011, 06:28 AMSorobanRe: what possible values of k.
Hello, rcs!

Quote:

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

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

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The problem is not clearly stated.

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

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

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