Question

if the ratio of roots of the quadratic equation x2+px+q=0 be equal to the ratio of x2+lx+m=0. Prove that p2m=l2q​

Answer 1

SOLUTION ☺️

$= > let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: equation \\ = > x {}^{2} + px + q = 0 \\ = > and \: a \: and \: b \: are \: the \: roots \: of \: equation \\ = > x {}^{2} + lx + m = 0 \\ = > then \: \alpha + \beta = - p \: and \: \alpha \beta = q.....(1) \\ = > a + b = - l \: and \: ab = m.....(2) \\ \\ given \\ = > \frac{ \alpha }{ \beta } = \frac{a}{b} .....(3) \\ = > \frac{ \beta }{ \alpha } = \frac{b}{a} ........(4) \\ = > adding \: (3) \: and \: (4) \: we \: get \\ = > \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{a}{b} + \frac{b}{a} \\ = > \frac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } + 2 = \frac{a {}^{2} + b {}^{2} }{ab} + 2 \\ = > \frac{ \alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta }{ \alpha \beta } = \frac{a {}^{2} + b {}^{2} + 2ab}{ab} \\ = > \frac{ (\alpha + \beta ) {}^{2} }{ \alpha \beta } = \frac{(a + b) {}^{2} }{ab} \\ = > \frac{ (- p) {}^{2} }{q} = \frac{ ( - l) {}^{2} }{m} \\ = > \frac{p {}^{2} }{q} = \frac{l {}^{2} }{m} \\ = > p {}^{2} m = l {}^{2} q \: \: hence \: proved$

HOPE it helps ✔️

Answer 2

We are given that

α / β = γ / δ (3)

Reciprocating both the sides, we'll get

β / α = δ / γ (4)

Adding (3) and (4), we'll get

=> (α / β) + (β / α) = (γ / δ) + (δ / γ)

=> (α^2 + β^2) /αβ = (γ^2 + δ^2) /γδ

Adding 2 on both the sides,

=> [(α^2 + β^2) /αβ] + 2 = [(γ^2 + δ^2) /γδ] + 2

=> (α^2 + β^2 + 2αβ) /αβ = (γ^2 + δ^2 + 2γδ) /γδ

=> (α + β)^2 /αβ = (γ + δ)^2 /γδ

Now, using α + β = -p, αβ = q, γ + δ = -r, γδ = m,

=> (-p)^2 /q = (-r)^2 /m

=> m.p^2 = q.r^2

Hence Proved.