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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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Sol. Let the zeroes be $\alpha$ and $6\alpha$
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
Rightarrow 7
alpha =
frac{14}{p}
Rightarrow
alpha =
frac{2}{p}$$\begin{aligned}& (1/2 Mark) \\ & Product of zeroes\end{aligned}$$=
alpha
times 6
alpha =
frac{8}{p}$$\begin{aligned}& (1/2 Mark) \\ & Rightarrow 6(\frac{2}{p})^2 = \frac{8}{p}\end{aligned}$$ (1/2 Mark)
Rightarrow p = 3$ (1 Mark)
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
Rightarrow 7
alpha =
frac{14}{p}
Rightarrow
alpha =
frac{2}{p}$$\begin{aligned}& (1/2 Mark) \\ & Product of zeroes\end{aligned}$$=
alpha
times 6
alpha =
frac{8}{p}$$\begin{aligned}& (1/2 Mark) \\ & Rightarrow 6(\frac{2}{p})^2 = \frac{8}{p}\end{aligned}$$ (1/2 Mark)
Rightarrow p = 3$ (1 Mark)