Question

if roots of (p^2+q^2)x^2 -2(pr+Q's)x+(r^2+s^2)=0 are equal prove p/q=r/s​

Answer 1

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

$\sf \: A \: quadratic \: equation \: ( {p}^{2} + {q}^{2}) {x}^{2} - 2(pr + qs)x + ( {r}^{2} + {s}^{2}) = 0$

$\sf \: has \: real \: and \: equal \: roots.$

$\large\underline{\sf{To\:prove - }}$

$\sf \: \dfrac{p}{q} = \dfrac{r}{s}$

$\large\underline{\sf{Basic \: Concept \: Used - }}$

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

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

The given Quadratic equation is

$\rm :\longmapsto\:( {p}^{2} + {q}^{2}) {x}^{2} - 2(pr + qs)x + ( {r}^{2} + {s}^{2}) = 0$

On comparing with ax² + bx + c = 0, we get

$\: \: \: \: \: \bull \sf \: a = {p}^{2} + {q}^{2}$

$\: \: \: \: \: \bull \sf \: b = - 2(pr + qs)$

$\: \: \: \: \: \bull \sf \: c = {r}^{2} + {s}^{2}$

Since,

• Given equation has real and equal roots,

So,

$\bf\implies \:Discriminant \: (D) = 0$

$\rm :\implies\: {b}^{2} \: - \: 4ac = 0$

$\rm :\longmapsto\: {\bigg( - 2(pr + qs) \bigg) }^{2} - 4({p}^{2} + {q}^{2} )({r}^{2} + {s}^{2} ) = 0$

$\rm :\longmapsto\: 4{\bigg((pr + qs) \bigg) }^{2} - 4({p}^{2} + {q}^{2} )({r}^{2} + {s}^{2} ) = 0$

On dividing by 4, we get

$\rm :\longmapsto\: {\bigg(pr + qs \bigg) }^{2} - ({p}^{2} + {q}^{2} )({r}^{2} + {s}^{2} ) = 0$

$\sf \: {p}^{2} {r}^{2} + {q}^{2} {s}^{2} + 2pqrs - ( {p}^{2} {r}^{2} + {p}^{2} {s}^{2} + {q}^{2} {r}^{2} + {q}^{2} {s}^{2} ) = 0$

$\sf \: \cancel{{p}^{2}{r}^{2}}+\cancel{{q}^{2}{s}^{2}}+2pqrs - \cancel{{p}^{2} {r}^{2}} - {p}^{2} {s}^{2} - {q}^{2} {r}^{2} - \cancel{{q}^{2}{s}^{2}} = 0$

$\rm :\longmapsto\:2pqrs - {p}^{2} {s}^{2} - {q}^{2} {r}^{2} = 0$

$\rm :\longmapsto\:{p}^{2} {s}^{2}+{q}^{2}{r}^{2} - 2pqrs = 0$

$\rm :\longmapsto\: {(ps)}^{2} + {(qr)}^{2} - 2 \times (ps) \times (qr) = 0$

$\rm :\implies\: {\bigg(ps \: - \: qr \bigg) }^{2} = 0$

$\rm :\implies\:ps \: - \: qr \: = \: 0$

$\rm :\longmapsto\:ps \: = \: qr$

$\bf\implies \:\dfrac{p}{q} \: = \: \dfrac{r}{s}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$