Question

If one root of the quadratic equation ${x^2} - x + 3a = 0$ is double the root of ${x^2} - x + a = 0$ . Find a ($a \ne 0$).

Answer 1

Hint: For a given quadratic equation, there are two roots say \[2\alpha \] and $\beta $. There is a second quadratic equation and it is given that the one root of the first quadratic equation is equal to double the root of the second quadratic equation. Therefore, one root of the second equation will be \[\alpha \].
The roots of the quadratic equation satisfy the equation i.e. if we put $x = 2\alpha $then it satisfies the equation. So, we replace \[x\] with one its roots in both equations let say in first equation replace \[x\] with \[2\alpha \] and in second equation replace \[x\] with \[\alpha \]. Now, we got the equations in terms of a and \[\alpha \]. By solving these equations, you can find the value of a.

Complete step-by-step answer:
Let \[2\alpha \] and $\beta $ be the roots of the quadratic equation given below:
$ \Rightarrow {x^2} - x + a = 0$ ………….(1)
If \[2\alpha \] is the root of above equation then $x = 2\alpha $ satisfy the equation and hence by replacing \[x\] with \[2\alpha \] in this equation we get,
$ \Rightarrow {(2\alpha )^2} - 2\alpha + 3a = 0$
By opening the bracket, we get,
$ \Rightarrow 4{\alpha ^2} - 2\alpha + 3a = 0$ ………….(2)
It is given that one of the roots of this equation is double the root of second equation i.e. given to us as follows:
$ \Rightarrow {x^2} - x + a = 0$ ………….(3)
If \[2\alpha \] is one of the roots of equation (1) then \[\alpha \] will be the root of equation (3) and $x = \alpha $ will satisfy equation (3). By replacing \[x\] with \[\alpha \] in the equation (3) we get,
$ \Rightarrow {\alpha ^2} - \alpha + a = 0$ ……………(4)
To obtain the value of a we will solve the equation (2) and (4) as follows:
By subtracting 3 times equation (4) from 3 times equation (2) We get,
\[ \Rightarrow 4{\alpha ^2} - 2\alpha + 3a - 3({\alpha ^2} - \alpha + a) = 0\]
By opening the bracket, we get,
\[ \Rightarrow 4{\alpha ^2} - 2\alpha + 3a - 3{\alpha ^2} + 3\alpha - 3a = 0\]
By combining like terms, we get,
\[ \Rightarrow 4{\alpha ^2} - 3{\alpha ^2} - 2\alpha + 3\alpha + 3a - 3a = 0\]
By adding or subtracting like terms we get,
\[ \Rightarrow {\alpha ^2} + \alpha = 0\]
By taking \[\alpha \] common from both terms we get,
\[ \Rightarrow \alpha (\alpha - 1) = 0\]
By equating both sided we get,
$ \Rightarrow \alpha = 0$ and $\alpha - 1 = 0$
$ \Rightarrow \alpha = 0$ and $\alpha = 1$
Now put $\alpha = 0$in equation (2) we get,
$ \Rightarrow {0^2} - 0 + 3a = 0$
$
   \Rightarrow 3a = 0 \\
   \Rightarrow a = 0 \\
$
As it is given $a \ne 0$ so this value in cancelled and put $\alpha = 1$ in equation (2) we get,
$
   \Rightarrow 4{(1)^2} - 2(1) + 3a = 0 \\
   \Rightarrow 4 - 2 + 3a = 0 \\
$
By solving and taking a one side and constant other side we get,
$
   \Rightarrow 3a = - 2 \\
   \Rightarrow a = \dfrac{{ - 2}}{3} \\
$
Hence, $a = \dfrac{{ - 2}}{3}$ is the required answer.

Note: In this student can get confused in taking roots i.e. for first equation they take \[\alpha \]and for second equation they take \[2\alpha \] but it is wrong as they have given the root of first equation is double the root of second equation. Take care of this.
During calculation while opening the bracket take care of the sign. Generally, students can make mistakes in calculation.
Never take the other root of both the equations into consideration as they haven’t asked for that. This can waste your time.