Question

If a,b are roots of equation

x²+px-q=0 and

c,d are roots of equation

x²+px+r=0

Prove that

(a-c)(a-d)=(b-c)(b-d)=q+r

Answer 1

Here is a slight difference in variables used,question is same
x^2 + px - q = 0 has roots α, β
x^2 + px + r = 0 has roots γ, δ

Using Sum of roots and Product of roots formulae on both these equations, we get these results-

α + β = -p (1)
αβ = -q (2)

γ + δ = -p (3)
γδ = r (4)


Lets have a look at what we have to prove.

(α - γ)(α - δ) = (β - γ)(β - δ) = q + r

We will begin with (α - γ)(α - δ).



=> (α - γ)(α - δ)
=> α^2 - αδ - αγ + γδ
=> α^2 - α(δ + γ) + γδ

From (3) and (1), it is understood that (α + β) = (γ + δ). We'll use it here,

=> α^2 - α(α + β) + γδ
=> α^2 - α^2 - αβ + γδ
=> - αβ + γδ

From (2) and (4), we know that αβ = -q and γδ = r. Using these here, we'll get

=> -(-q) + r
=> q + r <proved>

Similarly, we'll go with (β - γ)(β - δ)
=> (β - γ)(β - δ)
=> β^2 - βγ - βδ + γδ
=> β^2 - β(γ + δ) + γδ

Again, we'll use (α + β) = (γ + δ).

=> β^2 - β(α + β) + γδ
=> β^2 - αβ - β^2 + γδ
=> - αβ + γδ

Using αβ = -q and γδ = r again

=> -(-q) + r
=> q + r <proved>

So, we can say that (α - γ)(α - δ) = (β - γ)(β - δ) = q + r
Hence Proved.