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Find the value of p for …

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Find the value of p for which one root of the quadratic equation Px2 - 14x +8 is 6 times the other
  • 1 answers

Sia ? 6 years, 6 months ago

Let {tex}\alpha{/tex} and {tex}6\alpha{/tex} be the roots of equation.
We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8

Sum of zeroes{tex} = -\frac ba = -\frac{-14}{p}{/tex}

{tex}\alpha +6\alpha=\frac{14}{p}{/tex}
{tex}7\alpha = \frac{14}{p}{/tex}
{tex}\alpha = \frac2p{/tex}............(i)
Also, Product of the zeroes {tex} = \frac 8p=\frac ca{/tex}
{tex}\alpha \times 6\alpha = \frac 8p{/tex}
{tex}6\alpha^2=\frac 8p{/tex}
From (i)
{tex}6(\frac{2}{p})^2=\frac 8p{/tex}
{tex}6\times \frac {4}{p^2}=\frac 8p{/tex}
{tex}\frac{6}{p^2}=\frac2p{/tex}
{tex}\frac 62=\frac{p^2}{p}{/tex}
{tex}Hence, \ p=3{/tex}

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