Question

Prove that if both roots of x²+px+q = 0 are positive,
then the roots of the equations
qy²+(p-2rq)y+1-pr = 0 are positive for all r≥0 . Discuss case when r < 0

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Answer 1

$\large{\underline{\underline{\mathbf{Solution\:of\:Your\:Question\:!!}}}}$

Since, both roots of X² + px +q = 0 are positive

we must have :---

➜ p²-4q≥0, -p>0, q>0 .....................[a]

Now the discriminant of the second equation

$= {(p - 2rq)}^{2} - 4q(1 - pr)$

$4 {r}^{2} {q}^{2} + {p}^{2} - 4q \geqslant 0$

个个个个. _______ [p²-4q≥0{from a}]

Now, we must show that the sum and Product of the roots of the second equation are positive for all r ≥ 0.

$sum \: of \: the \: roots = - \frac{p - 2rq}{q} =\frac{2rq - p}{q}>0$

For all r ≥ 0 (because q>0, p<0 from (a))

and $product = \frac{1 - pr}{q} > 0 \: for \: all \: r \geqslant 0$

If r<0, then sum and product are positive provided the negative r is satisfying.

2rq-p > 0 and 1-pr > 0

i.e., $r = \frac{p}{2q}<strong>,</strong> r = \frac{1}{p}$

$\large\mathbb\red{Be\:BrainlY\:!!}$

Answer 2

Since, both roots of X² + px +q = 0 are positive

we must have :---

➜ p²-4q≥0, -p>0, q>0 .....................[a]

Now the discriminant of the second equation

= {(p - 2rq)}^{2} - 4q(1 - pr)=(p−2rq)

2

−4q(1−pr)

4 {r}^{2} {q}^{2} + {p}^{2} - 4q \geqslant 04r

2

q

2

+p

2

−4q⩾0

个个个个. _______ [p²-4q≥0{from a}]

Now, we must show that the sum and Product of the roots of the second equation are positive for all r ≥ 0.

sum \: of \: the \: roots = - \frac{p - 2rq}{q} =\frac{2rq - p}{q}>0sumoftheroots=−

q

p−2rq

=

q

2rq−p

>0

For all r ≥ 0 (because q>0, p<0 from (a))

and product = \frac{1 - pr}{q} > 0 \: for \: all \: r \geqslant 0product=

q

1−pr

>0forallr⩾0

If r<0, then sum and product are positive provided the negative r is satisfying.

2rq-p > 0 and 1-pr > 0

i.e., r = \frac{p}{2q}, r = \frac{1}{p}r=

2q

p

,r=

p

1