question_answer

Find the value of p, for which one root of the quadratic equation \[p{{x}^{2}}-14x+8=0\] is 6 times the other.

Answer:

Given, equation is \[p{{x}^{2}}-14x+8=0\]

Let                     one root \[=\alpha \],

then other root \[=6\alpha \]

Sum of roots \[=-\frac{b}{a}\];

\[\alpha +6\alpha =\frac{-(-14)}{p}\]

\[7\alpha =\frac{14}{p}\];

\[\alpha =\frac{14}{p\times 7}\]

Or                     \[\alpha =\frac{2}{p}\]                                                               ?(i)

Product of roots \[=\frac{c}{a}\]

\[(\alpha )(6\alpha )=\frac{8}{p}\]

\[6{{\alpha }^{2}}=\frac{8}{p}\]                                                                ?(ii)

Putting value of \[\alpha \] from eq. (i)

\[6{{\left( \frac{2}{p} \right)}^{2}}=\frac{8}{p}\]

\[\Rightarrow \]               \[6\times \frac{4}{{{p}^{2}}}=\frac{8}{p}\]

\[\Rightarrow \]               \[24p=8{{p}^{2}}\]

\[\Rightarrow \]               \[8p(p-3)=0\]

\[\Rightarrow \]                \[8{{p}^{2}}-24p=0\]

\[\Rightarrow \]               Either \[8p=0\Rightarrow p=0\]

Or                     \[p-3=0\Rightarrow p=3\]

For \[p=0\], given condition is not satisfied

\[\therefore \]                  \[p=3\]