Question

the sum of two roots of x²- px+q =0 is thrice of their difference.Then which is the correct relation_____
options
(a) 9p²=2q
(b) 2p²=9q
(c) 2p²+9q=0
(d) 9p²+2q=0
(please solve this sum)​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: zeroes \: of \: {x}^{2} - px + q \: be \: \alpha \: and \: \beta.$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \: \dfrac{ (- p)}{1} = p$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta = \dfrac{q}{1} = q$

Now,

According to statement,

$\rm :\longmapsto\: \alpha + \beta = 3( \alpha - \beta )$

$\rm :\longmapsto\:p = 3 \sqrt{ {( \alpha + \beta)}^{2} - 4 \alpha \beta }$

$\rm :\longmapsto\:p = 3 \sqrt{ {( p)}^{2} - 4q}$

On squaring both sides, we get

$\rm :\longmapsto\: {p}^{2} = 9( {p}^{2} - 4q)$

$\rm :\longmapsto\: {p}^{2} = 9{p}^{2} - 36q$

$\rm :\longmapsto\: 8{p}^{2} = 36q$

$\bf\implies \: {2p}^{2} = 9q$

Hence,

• Option (b) is correct.

Additional Information :-

$\sf \: Let \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d \: be \: \alpha, \: \beta \: and \: \gamma \: then$

$\rm :\longmapsto\: \alpha + \beta + \gamma = - \: \dfrac{b}{a}$

$\rm :\longmapsto\: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a}$

$\rm :\longmapsto\: \alpha \beta \gamma = - \: \dfrac{d}{a}$