Question

If the difference of the roots of x² + 2px + q = 0 and x² + 2qx + p = 0 is equal,then prove that
p+q + 1 = 0, where p#q.
​

Answer 1

Answer:

Proof

Step-by-step explanation:

We know that

(a+b)² - 2ab = a² + b²

(a-b)² + 2ab = a² + b²

Since RHS is equal their LHS will be equal

∴ (a-b)² = (a+b)² - 4ab  ------ (1)

Now let eq(A) = x² + 2px +  q = 0

             eq(B) = x² + 2qx +  p = 0

Let the roots of eq(A) be α,β

From eq(1) (α-β)² = (α+B)² - 4αβ

                             = (-2p)² - 4q       [∵ α + β = -b/a = -2p and αβ = c/a = q]

                             = √(4p² - 4q) ------ (2)

Let the roots of eq(B) be γ,δ

From eq(1) (γ-δ)² = (γ+δ)² - 4γδ

                             = (-2q)² - 4p       [∵ α + β = -b/a = -2q and αβ = c/a = p]

                             = √(4q² - 4p) ------ (3)

According to question eq(2) = eq(3)

∴ 4p² - 4q = 4q² - 4p

∴ p² - q² = -(p-q)

∴ (p+q)(p-q) = -(p-q)

∴ (p+q) = -1

∴ p+q+1 = 0

Hence proved