Question

If ${y^2} + 6y - 3m = 0$ and ${y^2} - 3y + m = 0$ have a common root, then find the possible values of m.
A. $0, - \dfrac{{27}}{{16}}$
B. $0, - \dfrac{{81}}{{16}}$
C. $0,\dfrac{{81}}{{16}}$
D. $0,\dfrac{{27}}{{16}}$

Answer 1

Hint: We will first find the value of m from the first equation given to us and then find the value of m by the second equation given to us and then compare those values to get the required answer.

Complete step-by-step solution:
We have the two equations as ${y^2} + 6y - 3m = 0$ …………….(1) and ${y^2} - 3y + m = 0$ ………………(2)
Since both the equations (1) and (2) have a common root.
Let that common root be ‘a’.
Since ‘a’ is the root of both the equations, it must satisfy both of the given equations ${y^2} + 6y - 3m = 0$ and ${y^2} - 3y + m = 0$.
Putting y = a in the equations (1), we will get the following equations:-
$ \Rightarrow {a^2} + 6a - 3m = 0$
Taking the ‘3m’ from subtraction in LHS to addition in RHS, we will then get:-
$ \Rightarrow 3m = {a^2} + 6a$ …………………(3)
Putting y = a in the equations (2), we will get the following equations:-
$ \Rightarrow {a^2} - 3a + m = 0$
Taking the ‘m’ from addition in LHS to subtraction in RHS, we will then get:-
$ \Rightarrow m = - {a^2} + 3a$ …………………(4)
Multiplying the equation (4) by 3, we will get:-
\[ \Rightarrow 3m = 3\left( {3a - {a^2}} \right)\]
Simplifying the calculations on the RHS, we will then get:-
$ \Rightarrow 3m = 9a - 3{a^2}$ …………………(5)
Comparing the equations (3) and (5), we can write as:-
$ \Rightarrow {a^2} + 6a = 9a - 3{a^2}$
Taking all the terms with ‘a’ on the LHS, we will then obtain:-
\[ \Rightarrow {a^2} + 3{a^2} + 6a - 9a = 0\]
Combining the like terms on the LHS, we will obtain:-
\[ \Rightarrow 4{a^2} - 3a = 0\]
Taking ‘a’ common from LHS, we will get:-
\[ \Rightarrow a\left( {4a - 3} \right) = 0\]
We get either $a = 0$ or $a = \dfrac{3}{4}$.
Now, we will get the values of m possible by putting these in (4).
Case 1: a = 0
$ \Rightarrow m = - {a^2} + 3a$
Putting a = 0 in above expression, we will get:-
$ \Rightarrow m = - {0^2} + 3 \times 0$
Simplifying the calculations, we will get:-
$ \Rightarrow $m = 0
Case 2: $a = \dfrac{3}{4}$
$ \Rightarrow m = - {a^2} + 3a$
Putting $a = \dfrac{3}{4}$in above expression, we will get:-
$ \Rightarrow m = - {\left( {\dfrac{3}{4}} \right)^2} + 3 \times \dfrac{3}{4}$
Simplifying the calculations, we will get:-
$ \Rightarrow m = - \dfrac{9}{{16}} + \dfrac{9}{4}$
Taking the LCM, we will then obtain:-
$ \Rightarrow m = \dfrac{{ - 9 + 36}}{{16}}$
Simplifying the calculations to obtain the following:-
$ \Rightarrow m = \dfrac{{27}}{{16}}$

$\therefore $ The correct option is (D).

Note: The students must note that they need to find both the possible values of m by putting in both the possible values of the common root ‘a’.
The students can also verify the values of a in between by putting them in the given equations and if they satisfy the equations, that means we are going in the right direction.