# Find the values of k so that the area of the triangle with vertices $( 1,\ -1) ,\ ( -4,\ 2k)$ and $( -k,\ -5)$ is 24 sq. units.

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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