Quadratic EquationHard
Question
Let α, β be the roots of the equation x2 - px + r = 0 and
, 2β be the roots of the equation x2 - qx + r = 0. Then the value of r is
, 2β be the roots of the equation x2 - qx + r = 0. Then the value of r isOptions
A.
(p - q)(2q - p)
(p - q)(2q - p)B.
(q - p)(2p - q)
(q - p)(2p - q)C.
(q - 2p)(2q - p)
(q - 2p)(2q - p)D.
(2p - q)(2q - p)
(2p - q)(2q - p)Solution
The equation x2 - px + r = 0 has roots (α, β) and the equation
x2 - qx + r = 0 has roots
.
⇒ r = αβ and α + β = p and
+ 2β = q
⇒ β =
and α = 
⇒ αβ = r =
(2q - p)(2p - q).
x2 - qx + r = 0 has roots
. ⇒ r = αβ and α + β = p and
+ 2β = q⇒ β =
and α = 
⇒ αβ = r =
(2q - p)(2p - q).Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
Complete set of real values of k for which the inequality ${kx}^{2} - kx - 1 < 0$ holds for any real $x$, satisfy...Let f(x) = a6x6 + a5x5 + a4x4 + ...... + a0 be a real polynomial such that a0 ∈ R+ and a0 + The equation f(x) = 0...If α, β are roots of the equation 2x2 − 35x + 2 = 0, then the value of (2α − 35)3. (2β &...If x is real, then the values of the expression are not -...The roots of the equation (x + 2)2 = 4 (x + 1) − 1 are -...