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Find the value of $p$, for which one zero of the quadratic polynomial $px^2-14x + 8$ is $6$ times the other.
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Let the zeroes be $\alpha$ and $6\alpha$
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
⇒ 7α = $\frac{14}{p}$ ⇒ α = $\frac{2}{p}$ (1/2 Mark)
Product of zeroes $= \alpha \times 6\alpha = $8/p (1/2 Mark)
⇒ 6(2/p)² = 8/p (1/2 Mark)
⇒ p = 3 (1 Mark)
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
⇒ 7α = $\frac{14}{p}$ ⇒ α = $\frac{2}{p}$ (1/2 Mark)
Product of zeroes $= \alpha \times 6\alpha = $8/p (1/2 Mark)
⇒ 6(2/p)² = 8/p (1/2 Mark)
⇒ p = 3 (1 Mark)