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Formation of a Quadratic Equation From It's Roots

State whether...

Question

State whether x = $\frac{- k}{2}$ is the root of the quadratic equation 2x² + (k – 6)x – 3k = 0

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Solution

Given: 2x² + (k – 6)x – 3k = 0 and x = $\frac{- k}{2}$ .
On substituting x = $\frac{- k}{2}$ in L.H.S. of the given equation, we get:

$2 {(\frac{- k}{2})}^{2} + (k - 6) (\frac{- k}{2}) - 3 k = \frac{2 k^{2}}{4} - \frac{k^{2}}{2} + 3 k - 3 k = \frac{2 k^{2} - 2 k^{2}}{4} = 0$

= >L.H.S. = R.H.S.
Thus, x = $\frac{- k}{2}$ satisfies the given equation.
Therefore, x = $\frac{- k}{2}$ is a root of the given quadratic equation.

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