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Question

Answers

A). $1,\dfrac{{ - 1}}{2}$

B). 2 or 4

C). 4

D). 3

Answer
Verified

Given , $(3m + 1){x^2} + 2(m + 1)x + m = 0$

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 $1,\dfrac{{ - 1}}{2}$.

(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.