Quadratic EquationEasy
Question
The value of α for which the sum of the squares of the roots of the equation x2 - (a - 2)x - a - 1 = 0 assume the least value is
Options
A.1
B.0
C.3
D.2
Solution
x2 - (a - 2)x - a - 1 = 0
⇒ α + β = a - 2
αβ = - (a + 1)
α2 + β2 = (α + β)2 - 2αβ
a2 - 2a + 6 = (a - 1)2 + 5
⇒ a = 1.
⇒ α + β = a - 2
αβ = - (a + 1)
α2 + β2 = (α + β)2 - 2αβ
a2 - 2a + 6 = (a - 1)2 + 5
⇒ a = 1.
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 the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{\text{ }b}$ be $\frac{5}{16},a > 2$. If $\alpha$ is such that ...The roots of the equation (a + b − 2c) x2 − (2a − b - c) x + (a − 2b+ c) = 0 are -...The roots of the equation (a2 + b2) x2 − 2(bc + ad) x + (c2 + d2) = 0 are equal, if -...If product of roots of the equation x2 − 4mx + 3e2 log m − 4 = 0 is 8, then its roots are real, when m equal...