If p,q,r,s belongs to R and pr=2(q+s) then show that atleast one of the eqn x²+px+q=0,x²+rx+s=0 has real roots.
To prove that at least one of the equations x² + px + q = 0 or x² + rx + s = 0 has real roots, we can approach the problem using contradiction.
Assume that both equations have complex roots, which means their discriminants are negative.
For the first equation, x² + px + q = 0, the discriminant is given by D₁ = p² - 4q.
Similarly, for the second equation, x² + rx + s = 0, the discriminant is given by D₂ = r² - 4s.
Since both discriminants D₁ and D₂ are negative, we have:
D₁ < 0 and D₂ < 0
Expanding D₁ and D₂ using the given equation pr = 2(q + s), we get:
p² - 4q < 0 and r² - 4s < 0
Adding both inequalities, we have:
p² - 4q + r² - 4s < 0
Rearranging the terms, we get:
(p² + r²) - 4(q + s) < 0
Substituting the given equation pr = 2(q + s), we have:
(p² + r²) - 4(pr/2) < 0
Simplifying further:
p² + r² - 2pr < 0
Notice that the left side of the inequality represents (p - r)², which must be greater than or equal to zero. Therefore, we have:
(p - r)² < 0
However, a squared term can never be negative, which contradicts our assumption that both equations have complex roots.
Hence, at least one of the equations x² + px + q = 0 or x² + rx + s = 0 must have real roots.