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Find the value of p, for which one root of the quadratic equation $px^{2} -14x+8=0$ is 6 times the other.
Given: Quadratic equation $px^{2} -14x+8=0$
To do: To Find the value of p, for which one root of the given quadratic equation is 6 times the other.
Solution:
Given Quadratic Equation\ px^{2} -14x+8=0\ $
on dividing bnoth sides from $p.$
$x^{2} -\frac{14}{p} x+\frac{8}{p} =0$
Also, one root is 6 times the other
Lets say one root $=x$
Second root $=6x$
as known : in a quadratic equation $ax^{2} +bx+c=0$
Sum of the roots $( \alpha +\beta ) =-b$
product of the roots $( \alpha \beta ) =c$
From the given equation:
Sum of the roots $=-\left( -\frac{14}{p}\right) =+\frac{14}{p}$
$\Rightarrow x+6x=\frac{14}{p}$
$\Rightarrow 7x=\frac{14}{p}$
$\Rightarrow x=\frac{2}{p} \ \ \ \ \ \ \ \ ..............( 1)$
Product of root
$x( 6x) =\frac{8}{p}$
$\Rightarrow 6x^{2} =\frac{8}{p}$
$\Rightarrow 6\left(\frac{2}{p}\right)^{2} =\frac{8}{p}$
$\Rightarrow \frac{24}{p^{2}} =\frac{8}{p}$
$\Rightarrow \frac{3}{p} =1$
$\Rightarrow p=3$
Thus for $p=3$, The given quadratic equation will have two roots such that one root is 6 times the another.
Updated on: 10-Oct-2022
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