Find the value of p, for which one Root of the equation pxsquare- 14x+18=0is 6 times the other

Find the value of p, for which one Root of the equation pxsquare- 14x+

1.	Find the value of p, for which one Root of the equation pxsquare- 14x+18=0is 6 times the other
Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}6\\alpha{/tex}\xa0be the roots of equation.We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8Sum 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\xa0{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}