Find the value of p, for which one root of the quadratic equation `px^(2)-14x+8=0` is 6 times the other.

Explanation

Let `alphaandbeta` be two roots of quadratic equation `px^(2)=14x+8=0`, such that
`beta=6alpha`
`:."Sum of roots"=(-b)/(a)`
`impliesalpha+6alpha=-((-14))/(p)`
`implies7alpha-(14)/(p)impliesalpha=(2)/(p)`
Product of roots `-(c)/(a)`
`impliesalpha.6alpha=(8)/(p)`
`implies6alpha^(2)=(8)/(p)impliesalpha^(2)=(8)/(6p)`
From equations (1) and (2), we get
`((2)/(p))^(2)=(8)/(6p)implies(4)/(p^(2))=(4)/(3p)`
`impliesp^(2)=3p`
`impliesp^(2)-3p=0" "("don't cancel p both sides")`
`impliesp(p-3)=0`
:.Either p=0 or p=3.
But p=0 is not possible, as on putting, p=0 in the given equation, we don't have a quatratic equation and therefore we cannot get two roots.
Hence, p=3
Alternatively,
Let one root of quadratic equation `px^(2)-14x+8=0` is `alpha`.
`:.palpha^(2)-14alpha+8=0" ".....(I)`
:. Other root of equation will be `6alpha`.
`:.p(6alpha^(2))-14(6alpha)+8=0`
`implies36palpha^(2)-84alpha+8=0`
`implies9palpha^(2)-21alpha+2=0" "......(2)`
Solving equations (1) and (2) by cross-multiplication method.
`:.(alpha^(2))/(-14(2)-8(-21))=(alpha)/(8(9p)-2p)=(1)/(p(-21)-9p(-14))`
`:.(alpha^(2))/(-28+168)=(alpha)/(70p)=(1)/(105p)`
`:.(alpha^(2))/(140)=(1)/(105p)and(alpha)/(70p)=(1)/(105p)`
`implies alpha^(2)=(140)/(105p)=(4)/(3p)andalpha=(70p)/(105p)=(2)/(3)`
`:.((2)/(3))^(2)=(4)/(3p)implies(4)/(9)=(4)/(3p)implies3p=9`
`:.p=3`