Question

Question:-

Prove that the equation x^2(p^2+q^2)+2x(pr+qs)+r^2+s^2=0 has no real roots. If ps=qr , then show that the roots are real and equal.

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subject : maths

Class : 10
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please give correct answers​

Answer 1

Step-by-step explanation:

The given equation is x

2

−2px+p

2

−q

2

+2qr−r

2

=0

The roots of the given equation are rational only when the discriminant is a perfect square.

The discriminant of the given equation is b

2

−4ac=(2p)

2

−4(p

2

−q

2

+2qr−r

2

)

⇒b

2

−4ac=4p

2

−4p

2

+4(q

2

−2qr+r

2

)

⇒b

2

−4ac=4(q−r)

2

=(2q−2r)

2

Here we can see that b

2

−4ac is a perfect square.

Therefore the roots of given equation are rational.

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Answer 2

Answer:-

given:-

${x}^{2}( {p}^{2} + {q}^{2}) + 2x(pr + qs) + {r}^{2} + {s}^{2} = 0$

Solve:-

$a = {p}^{2} + {q}^{2} .b = 2(pr + qs).c = {r}^{2} + {s}^{2}$

$\triangle = {b}^{2} - 4ac = [2(pr + qs)] ^{2} - 4( {p}^{2} + {q}^{2} )( {r}^{2} + {s}^{2})$

$= 4 [ {p}^{2} {r}^{2} + 2pqrs + {q}^{2} {s}^{2} - {p}^{2} {r}^{2} - {p}^{2} {s}^{2} - {q}^{2} {r}^{2} - {q}^{2} {s}^{2} ]$

$= 4 [ - {p}^{2} {s}^{2} + 2pqrs - {q}^{2} {r}^{2} ] = - 4 [(ps - qr)^{2} < 0]$

$\triangle = {b}^{2} - 4ac < 0.$

$if.ps = qr \: then \: \triangle = - 4 [ps - qr]^{2} = - 4[qr - qr] ^{2} = 0$