If the ratio of roots of the equation x^2 + px + q = 0 is equal to the ratio of roots of x^2 + rx + m = 0, prove that m.p^2 = q.r^2

First of all, lets have a look at what we're given. Ratio of roots of x^2 + px + q = 0 and x^2 + rx + m = 0 are equal.

We are assuming that the roots of x^2 + px + q = 0 are α, β while roots of x^2 + rx + m = 0 are γ, δ. So, as we're given,

α / β = γ / δ

By using the sum and product of roots formulae, we can say that

α + β = -p ; αβ = q
γ + δ = -r ; γδ = m

We have to prove that m.p^2 = q.r^2. Now, there can be several approaches to the proof. We'll be telling you two of them.

APPROACH 1 - To begin with Left Hand Side,
LHS
=m.p^2
= (γδ) [-(α + β)]^2               (by using γδ = m and α + β = -p)
= (γδ) (α^2 + β^2 + 2αβ)
= γδ.α^2 + γδ.β^2 + 2αβγδ

Lets have a look at the Right Hand Side now,
RHS
= q.r^2
= (αβ) [-(γ + δ)]^2               (by using αβ = q and γ + δ = -r)
= (αβ) (γ^2 + δ^2 + 2γδ)
= αβ.γ^2 + αβ.δ^2 + 2αβγδ

One can easily notice that the expressions LHS and RHS have lead us to look very similar.
LHS = γδ.α^2 + γδ.β^2 + 2αβγδ
RHS = αβ.γ^2 + αβ.δ^2 + 2αβγδ

Both LHS and RHS already have their third terms same. We only have to prove the first two terms of LHS equal to that of RHS, which might look a tough job but it actually isn't. We only have to follow the basic condition given in the question, i.e. ratio of roots of both the equations are equal, or in mathematical terms,
α / β = γ / δ
=> αδ = βγ      (on cross multiplication)

Lets take the first term of LHS
γδ.α^2
= γδαα

Notice the δα in the middle, which can be replaced with βγ, as we know that αδ = βγ.

= γδαα
= γβγα
= αβ.γ^2               (1)
which is the first term of RHS. Yes, Its that simple! You only have to look for the right thing so that you can use it to make something useful of it. The same can be done with second term of LHS.

γδ.β^2
= γδββ
= αδδβ
= αβ.δ^2               (2)

From (1) and (2), we can say that

LHS
= γδ.α^2 + γδ.β^2 + 2αβγδ
= αβ.γ^2 + αβ.δ^2 + 2αβγδ
= RHS

Hence Proved.

APPROACH - 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.