Solveeit Logo
Login

Question

Question: If the sum of the square of the roots of the equation \[{x^2} - \left( {a - 2} \right)x - a + 1 = 0\...

If the sum of the square of the roots of the equation x2−(a−2)x−a+1=0{x^2} - \left( {a - 2} \right)x - a + 1 = 0 is least, then find the value of a.
A)−1 - 1
B)11
C)22
D)−2 - 2

Explanation

Solution

We know that the sum of the roots =−coefficient of xcoefficient of x2 - \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}} and The product of the roots= constantcoefficient of x2\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}} . Use the formula of (a+b)2=a2+b2+2ab{\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 x2−(a−2)x−a+1=0{x^2} - \left( {a - 2} \right)x - a + 1 = 0--- (i)
Let us assume the roots of the given equation to be α\alpha and β\beta . We know that –
Sum of the roots=−coefficient of xcoefficient of x2 - \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}
And Product of the roots=constantcoefficient of x2\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}
On putting the given values we get,
⇒α+β=−−(a−2)1=(a−2)\Rightarrow \alpha + \beta = - \dfrac{{ - \left( {a - 2} \right)}}{1} = \left( {a - 2} \right) and
⇒αβ=−(a+1)1=−(a+1)\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-(a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
On putting the given values we get,
⇒(α+β)2=α2+β2+2αβ\Rightarrow {\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta
We can write it as-
⇒α2+β2=(α+β)2−2αβ\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,
⇒α2+β2=(a−2)2−2[−(a+1)]\Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} - 2\left[ { - \left( {a + 1} \right)} \right]
On opening the squares we get,
⇒α2+β2=(a2+4−4a)2+2a+2\Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {{a^2} + 4 - 4a} \right)^2} + 2a + 2
On simplifying we get,
⇒α2+β2=a2−4a+2a+4+2\Rightarrow {\alpha ^2} + {\beta ^2} = {a^2} - 4a + 2a + 4 + 2
⇒α2+β2=a2−2a+6\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-
⇒a2−2a+1+5\Rightarrow {a^2} - 2a + 1 + 5
We can write (a−1)2=a2+2a+1{\left( {a - 1} \right)^2} = {a^2} + 2a + 1 sop we get,
⇒(a−1)2+5\Rightarrow {\left( {a - 1} \right)^2} + 5
For this value to be least (a−1)2=0{\left( {a - 1} \right)^2} = 0
On solving this we get,
⇒a−1=0\Rightarrow a - 1 = 0
⇒a=1\Rightarrow a = 1
So the value of a is 11
Hence correct answer is ‘B’.

Note: You can also use the following formula after eq. (ii) as the coefficient of a2=1>0{a^2} = 1 > 0 for finding the least positive value of a,
⇒a=−12coefficient of acoefficient of a2\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,
⇒a=−12×−21\Rightarrow a = - \dfrac{1}{2} \times \dfrac{{ - 2}}{1}
On solving we get,
⇒a=−(−2)2\Rightarrow a = \dfrac{{ - \left( { - 2} \right)}}{2}
⇒a=22=1\Rightarrow a = \dfrac{2}{2} = 1
So we can also find the value of using this formula.