Question

If p , q , r and s are real numbers such that pr = 2 (q + s) , then show that at least one of the equations x² + px + q = 0 and x² + rx + s = 0 , has real roots

Answer 1

hope this helps you......

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, we have

x² + px + q = 0   ..... (i)

x² + rx + s = 0 ..... (ii)

Let D₁ and D₂ be the discriminant of equations (i) and (ii).

Then,

D₁ = p² - 4q and D₂ = r² - 4s

Adding both sides, we get

⇒ D₁ + D₂ = p² - 4q + r² - rs

⇒ D₁ + D₂ = (p² + r²) - 4(q + s)

⇒ D₁ + D₂ = p² + r² - 4(pr/2)

⇒ D₁ + D₂ = p² + r² - 2pr

⇒ D₁ + D₂ = (p - r)² ≥ 0

At least one of D₁ and D₂ is greater than or equal to zero.

At least one of the two equations has real roots.