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