Question

The number of quadratic equations which are unchanged by squaring their roots are
A.2
B.4
C.6
D.8

Answer 1

Hint: We can take $\alpha $ and $\beta $ as the roots of the quadratic equations. Then we can square the roots. Then we can equate the sum of the roots and the product of the roots before and after squaring. We can solve the equation of products of the root and find all the possible sets of values $\alpha $ and $\beta $ can take. Each pair of $\alpha $ and $\beta $ can form an equation. So the number of sets of $\alpha $ and $\beta $ will be the number of equations.

Complete step-by-step answer:
Let us assume $\alpha $ and $\beta $ are the roots of the quadratic equations.
It is given that the roots are squared. Then, ${\alpha ^2}$ and ${\beta ^2}$ are the roots
As the equation is not changing, the sum of the roots in both cases will be equal.
 $ \Rightarrow \alpha + \beta = {\alpha ^2} + {\beta ^2}$ .. (1)
We can take the product of the roots and equate them.
 $ \Rightarrow \alpha \beta = {\alpha ^2}{\beta ^2}$
On rearranging, we get,
 $ \Rightarrow {\alpha ^2}{\beta ^2} - \alpha \beta = 0$
On taking the common factors, we get,
 $ \Rightarrow \alpha \beta \left( {\alpha \beta - 1} \right) = 0$
This implies that , $\alpha = 0$ or $\beta = 0$ and $\alpha \beta = 1$
When $\alpha = 0$ ,
On substituting in (1),
 $ \Rightarrow 0 + \beta = 0 + {\beta ^2}$
On simplification we get,
  \[ \Rightarrow {\beta ^2} - \beta = 0\]
On taking $\beta $ common we get,
 \[ \Rightarrow \beta \left( {\beta - 1} \right) = 0\]
On simplification we get,
   \[ \Rightarrow \beta = 0,\beta = 1\]
So, the roots will be, 0,0 and 0,1.
For $\beta = 0$ , we get the same roots of 0,0 and 0,1
When, $\alpha \beta = 1$ ,
Equation (1) can be written as,
Adding and subtracting $2\alpha \beta $ from LHS we get,
 $ \Rightarrow \alpha + \beta = {\alpha ^2} + {\beta ^2} + 2\alpha \beta - 2\alpha \beta $
As \[{a^2} + 2ab + {b^2} = {(a + b)^2}\] , using this we get,
 $ \Rightarrow \alpha + \beta = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $
On substituting, $\alpha \beta = 1$ , we get,
 $ \Rightarrow \alpha + \beta = {\left( {\alpha + \beta } \right)^2} - 2$
We can make a substitution $t = \alpha + \beta $
 $ \Rightarrow t = {t^2} - 2$
On rearranging, we get,
 $ \Rightarrow {t^2} - t - 2 = 0$
On factorising, we get,
 $ \Rightarrow {t^2} - 2t + t - 2 = 0$
Taking terms common we get,
 $ \Rightarrow t\left( {t - 2} \right) + \left( {t - 2} \right) = 0$
Taking t-2, we get,
 $ \Rightarrow \left( {t - 2} \right)\left( {t + 1} \right) = 0$
So, we get, $t = \alpha + \beta = 2, - 1$
When $\alpha + \beta = 2,$ and $\alpha \beta = 1$ ,
The roots are 1,1.
When $\alpha + \beta = - 1$ and $\alpha \beta = 1$ ,
The equation will be ${x^2} + x + 1$ and it will have the roots $\omega ,{\omega ^2}$
Now we have 4 sets of roots,
 $\left( {0,0} \right)$ , $\left( {0,1} \right)$ , $\left( {1,1} \right)$ and $\left( {\omega ,{\omega ^2}} \right)$
Thus, we can have 4 equations which are unchanged by squaring the roots.
So, the correct answer is option B.

Note: We must not mix up the values of $\alpha $ and $\beta $. We can write them as pairs. The set of solutions when $\alpha = 0$ or $\beta = 0$ are the same. So, we only consider any one of them. The terms $1,\omega ,{\omega ^2}$ are known as cube roots of unity and they are the solution of the equation. Its values are given by $\omega = \dfrac{{ - 1}}{2} - i\dfrac{{\sqrt 3 }}{2}$ and ${\omega ^2} = \dfrac{{ - 1}}{2} + i\dfrac{{\sqrt 3 }}{2}$ . The product of $\omega ,{\omega ^2}$ is given by $\omega \times {\omega ^2} = {\omega ^3} = 1$ .