The value of $$a$$ for which the sum of the squares of the roots of the equation $$2{x^2} - 2\left( {a - 2} \right)x - \left( {a + 1} \right) = 0$$ is least, is
A.
$$1$$
B.
$$\frac{3}{2}$$
C.
$$2$$
D.
None
Answer :
$$\frac{3}{2}$$
Solution :
If $$\alpha ,\beta $$ be the roots of the equation then $$\alpha + \beta = a - 2,\alpha \beta = - \frac{{a + 1}}{2}$$
Sum of square of roots
$$\eqalign{
& S = {\alpha ^2} + {\beta ^2} = \left( {\alpha + {\beta ^2}} \right) - 2\alpha \beta \cr
& = {\left( {a - 2} \right)^2} + \left( {a + 1} \right) = {a^2} - 3a + 5 \cr
& S = {a^2} - 3a + \frac{9}{4} + \frac{{11}}{4} \cr
& S = {\left( {a - \frac{3}{2}} \right)^2} + \frac{{11}}{4} \cr} $$
Clearly $$S$$ is least when
$$\eqalign{
& a - \frac{3}{2} = 0 \cr
& \Rightarrow a = \frac{3}{2} \cr} $$
Releted MCQ Question on Algebra >> Quadratic Equation
Releted Question 1
If $$\ell ,m,n$$ are real, $$\ell \ne m,$$ then the roots by the equation: $$\left( {\ell - m} \right){x^2} - 5\left( {\ell + m} \right)x - 2\left( {\ell - m} \right) = 0$$ are