Quadratic EquationHard
Question
If α and β are the roots of x2 + px + q = 0 and α4, β4 are the root of x2 - rx + s = 0,then the equation x2 - 4qx + 2q2 - r = 0, has always
Options
A.two real roots
B.two positive roots
C.two negayive rppts
D.one positive and one negative root
Solution
Since α, β are the roots of x2 + px + q = 0 and α4, β4 are the roots of x2 - rx + s = 0.
α + β = - p; α β = q; α4, β4 = r and α4, β4 = s
Let roots of x2 - 4qx + (2q2 - r) = 0 be α′ and β′
Now, α′β′ = (2q2 - r)
= (αβ)2 - (α4 + β4)
= -(α2 - β2)2 > 0
⇒ Roots are real and of opposite sign.
α + β = - p; α β = q; α4, β4 = r and α4, β4 = s
Let roots of x2 - 4qx + (2q2 - r) = 0 be α′ and β′
Now, α′β′ = (2q2 - r)
= (αβ)2 - (α4 + β4)
= -(α2 - β2)2 > 0
⇒ Roots are real and of opposite sign.
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