Roots of the equation $p x (2 - x) = x - 4 m$ $px(2-x)=x-4m$ are $\frac{a}{3}$ $a/3$ an $\frac{b}{3}$ $b/3$ and while $a + b = 9$ $a+b=9$ , $a b = 12$ $ab=12$ . Find $p$ $p$ and $m$ $m$ ?

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Roots of the equation $p x (2 - x) = x - 4 m$ $px(2-x)=x-4m$ are $\frac{a}{3}$ $a/3$ an $\frac{b}{3}$ $b/3$ and while $a + b = 9$ $a+b=9$ , $a b = 12$ $ab=12$ . Find $p$ $p$ and $m$ $m$ ?

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Answer 1 · 2017-02-21T09:48:29

$p = - 1$ $p=-1$ and $m = \frac{1}{3}$ $m=1/3$

Explanation:

In a quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ,

we have sum of roots as $- \frac{b}{a}$ $-b/a$ and product of roots as $\frac{c}{a}$ $c/a$

Hence, as $p x (2 - x) = x - 4 m$ $px(2-x)=x-4m$

we have $- p x^{2} + 2 p x - x + 4 m = 0$ $-px^2+2px-x+4m=0$

or $p x^{2} - (2 p - 1) x - 4 m = 0$ $px^2-(2p-1)x-4m=0$

and as its roots are $\frac{a}{3}$ $a/3$ and $\frac{b}{3}$ $b/3$ , we have

$\frac{a}{3} + \frac{b}{3} = \frac{2 p - 1}{p}$ $a/3+b/3=(2p-1)/p$ or $\frac{2 p - 1}{p} = \frac{a + b}{3} = \frac{9}{3} = 3$ $(2p-1)/p=(a+b)/3=9/3=3$

Hence $2 p - 1 = 3 p$ $2p-1=3p$ or $p = - 1$ $p=-1$

Further $\frac{a}{3} \times \frac{b}{3} = \frac{- 4 m}{p} = 4 m$ $a/3xxb/3=(-4m)/p=4m$ (as $p = - 1$ $p=-1$ )

Hence $4 m = \frac{a b}{9} = \frac{12}{9} = \frac{4}{3}$ $4m=(ab)/9=12/9=4/3$ i.e. $m = \frac{1}{3}$ $m=1/3$

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Roots of the equation $p x (2 - x) = x - 4 m$ $px(2-x)=x-4m$ are $\frac{a}{3}$ $a/3$ an $\frac{b}{3}$ $b/3$ and while $a + b = 9$ $a+b=9$ , $a b = 12$ $ab=12$ . Find $p$ $p$ and $m$ $m$ ?

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Roots of the equation px(2−x)=x−4mpx(2-x)=x-4m are a3a/3 an b3b/3 and while a+b=9a+b=9, ab=12ab=12. Find pp and mm?

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Roots of the equation $p x (2 - x) = x - 4 m$ $px(2-x)=x-4m$ are $\frac{a}{3}$ $a/3$ an $\frac{b}{3}$ $b/3$ and while $a + b = 9$ $a+b=9$ , $a b = 12$ $ab=12$ . Find $p$ $p$ and $m$ $m$ ?