Question

The number of quadratic equation which are unchanged by squaring their roots, is:
A.2
B.4
C.6
D.None of these

Answer 1

Hint: Let the roots of the quadratic equation be \[\alpha ,\beta \] . Now, on squaring the roots, the roots of the new quadratic equation is \[{\alpha ^2},{\beta ^2}\] . Use the relation of quadratic equation \[a{x^2} + bx + c = 0\] as \[\alpha + \beta = - \dfrac{b}{a}\] and \[\alpha \beta = \dfrac{c}{a}\] . Hence, solve the equation obtained using the above equation and hence calculate all possible ordered pairs of required quantities if the root of the equation is known then the equation formed from the new roots can be given as \[(x - \alpha )(x - \beta ) = 0\] .

Complete step-by-step answer:
So, we have to form the quadratic equation whose root on squaring their roots does not change.
So, let the quadratic equation’s roots be \[\alpha ,\beta \] .
Hence, the equation on squaring the root should not change. So, all the relations for both the roots of the quadratic equation should not change. Thus, following equation can be formed as
 \[\alpha + \beta = {\alpha ^2} + {\beta ^2}\] and another is \[\alpha \beta = {\alpha ^2}{\beta ^2}\] .
So, from the equation with product value of roots we can consider that
\[ \Rightarrow \] \[{\alpha ^2}{\beta ^2} - \alpha \beta = 0\]
Taking common and solving the above equation we get
\[ \Rightarrow \] \[\alpha \beta \left( {\alpha \beta - 1} \right) = 0\]
On further simplification
\[ \Rightarrow \] \[\alpha \beta = 0,1\]
Hence, either \[\alpha \beta = 0\] so possible ordered pair for \[\left( {\alpha ,\beta } \right) = \left( {0,0} \right)\] and also \[\left( {\alpha ,\beta } \right) = \left( {1,0} \right)\] or \[\left( {0,1} \right)\]
While for \[\alpha \beta = 1\] the possible ordered pairs are \[\left( {\alpha ,\beta } \right) = \left( {1,1} \right)\] or either \[\left( {\alpha ,\beta } \right) = \left( {\omega ,{\omega ^2}} \right)\] as \[{\omega ^3} = 1\] .
Hence, for all the other corresponding values we will see that the above equation of relation of roots will not be satisfied and so there are a total four quadratic equations possible.
Hence, option (B) is our correct answer.

Note: The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
The product of the roots of a quadratic equation is equal to the constant term (the third term),
divided by the leading coefficient.
Hence, use the relation and think for all ordered pairs satisfying both the equations correctly.
Also calculate and form the equation carefully without making mistakes. Also think about the ordered pair and also confirm them whether they are satisfying or not. Hence, calculate and select properly.