Question

If the sum of the square of the roots of the equation \[{x^2} - \left( {a - 2} \right)x - a + 1 = 0\] is least, then find the value of a.
A)$ - 1$
B)$1$
C)$2$
D)$ - 2$

Answer 1

Hint: We know that the sum of the roots =$ - \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$ and The product of the roots= $\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$ . Use the formula of ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to find the sum of the square of the roots, and by the given condition we can find the value of a.

Complete step by step answer:

Given equation is \[{x^2} - \left( {a - 2} \right)x - a + 1 = 0\]--- (i)
Let us assume the roots of the given equation to be $\alpha $ and $\beta $. We know that –
Sum of the roots=$ - \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
And Product of the roots=$\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
On putting the given values we get,
$ \Rightarrow \alpha + \beta = - \dfrac{{ - \left( {a - 2} \right)}}{1} = \left( {a - 2} \right)$ and
$ \Rightarrow \alpha \beta = \dfrac{{ - \left( {a + 1} \right)}}{1} = - \left( {a + 1} \right)$
Now it is given that the sum of the squares of the roots of given equation is least . So we can find the sum of the squares of the roots using the formula-${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
On putting the given values we get,
$ \Rightarrow {\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta $
We can write it as-
$ \Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $
We already know the value of the sum and product of the roots. So on putting their values in the formula we get,
\[ \Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} - 2\left[ { - \left( {a + 1} \right)} \right]\]
On opening the squares we get,
\[ \Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {{a^2} + 4 - 4a} \right)^2} + 2a + 2\]
On simplifying we get,
\[ \Rightarrow {\alpha ^2} + {\beta ^2} = {a^2} - 4a + 2a + 4 + 2\]
$ \Rightarrow {\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$ --- (ii)
We have to find the value of a for which this value will be least.
So we can write-
$ \Rightarrow {a^2} - 2a + 1 + 5$
We can write ${\left( {a - 1} \right)^2} = {a^2} + 2a + 1$ sop we get,
$ \Rightarrow {\left( {a - 1} \right)^2} + 5$
For this value to be least \[{\left( {a - 1} \right)^2} = 0\]
On solving this we get,
$ \Rightarrow a - 1 = 0$
$ \Rightarrow a = 1$
So the value of a is $1$
Hence correct answer is ‘B’.

Note: You can also use the following formula after eq. (ii) as the coefficient of ${a^2} = 1 > 0$ for finding the least positive value of a,
$ \Rightarrow a = - \dfrac{1}{2}\dfrac{{{\text{coefficient of a}}}}{{{\text{coefficient of }}{{\text{a}}^2}}}$
On putting the value of the coefficients of in the formula we get,
$ \Rightarrow a = - \dfrac{1}{2} \times \dfrac{{ - 2}}{1}$
On solving we get,
$ \Rightarrow a = \dfrac{{ - \left( { - 2} \right)}}{2}$
$ \Rightarrow a = \dfrac{2}{2} = 1$
So we can also find the value of using this formula.