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For what value of ‘m’ , the equation $(3m + 1){x^2} + 2(m + 1)x + m = 0$ have equal root?
A). $1,\dfrac{{ - 1}}{2}$
B). 2 or 4
C). 4
D). 3


Answer Verified Verified
Hint: Hint:- Start by comparing with the standard quadratic equation and find out the discriminant value, and according to the condition of equal roots equate D to zero i.e. D=0, Find out the required value by solving the relation.
Given , $(3m + 1){x^2} + 2(m + 1)x + m = 0$

Complete Step by step solution:
We know for any quadratic equation $A{x^2} + Bx + C = 0$, equal roots are possible only when discriminant(D) = 0. Which can be found by the formula $D = {B^2} - 4AC$.
On comparing with equation 1, we get
$ A = (3m + 1) \\
 B = 2(m + 1) \\
 C = m $
Using the above-mentioned concept we need to find discriminant (D), we get :
$ D = {\left[ {2(m + 1)} \right]^2} - 4 \times (3m + 1) \times (m) \\
   = 4({m^2} + 1 + 2m) - 4(3{m^2} + m) $
Here, we used the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$
Now , $D = 0$
$ \Rightarrow 4({m^2} + 1 + 2m) - 4(3{m^2} + m) = 0 \\
   \Rightarrow ({m^2} + 1 + 2m) = (3{m^2} + m) \\
   \Rightarrow {m^2} + 1 + 2m = 3{m^2} + m \\
   \Rightarrow 2{m^2} - m - 1 = 0 $
Splitting the middle term, we get
$ 2{m^2} - 2m + m - 1 = 0 \\
   \Rightarrow 2m(m - 1) + 1(m - 1) = 0 \\
   \Rightarrow (2m + 1)(m - 1) = 0 \\
   \Rightarrow 2m + 1 = 0,m - 1 = 0 \\
   \Rightarrow m = 1,\dfrac{{ - 1}}{2} $
Therefore, the values of m are $1,\dfrac{{ - 1}}{2}$.
So, option A is the correct answer.

Note: For any quadratic equation $A{x^2} + Bx + C = 0$, there are three types of roots available which can only be determined after calculating Discriminant(D) by the formula or relation $D = {B^2} - 4AC$ . Now the three types of roots are as follows:
(i). D>0 , Distinct and real roots exist and the roots are $\alpha = \dfrac{{ - B + \sqrt D }}{{2A}},\beta = \dfrac{{ - B - \sqrt D }}{{2A}}$
(ii). D=0, Real and equal roots exist and the roots are $\alpha = \beta = \dfrac{{ - B}}{{2A}}$
(iii). D<0, Imaginary roots exist.
For imaginary roots, there’s a whole different chapter and concept known as “Complex numbers and roots”.
All the above-mentioned formulas and concepts are very important from the school and competitive exam point of view. Therefore, it is recommended to practice them thoroughly. They are easily scorable chapters.