Solveeit Logo

Question

Question: For what value of ‘m’ , the equation \((3m + 1){x^2} + 2(m + 1)x + m = 0\) have equal root? A). \(...

For what value of ‘m’ , the equation (3m+1)x2+2(m+1)x+m=0(3m + 1){x^2} + 2(m + 1)x + m = 0 have equal root?
A). 1,121,\dfrac{{ - 1}}{2}
B). 2 or 4
C). 4
D). 3

Explanation

Solution

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)x2+2(m+1)x+m=0(3m + 1){x^2} + 2(m + 1)x + m = 0

Complete Step by step solution:
We know for any quadratic equation Ax2+Bx+C=0A{x^2} + Bx + C = 0, equal roots are possible only when discriminant(D) = 0. Which can be found by the formula D=B24ACD = {B^2} - 4AC.
On comparing with equation 1, we get
A=(3m+1) B=2(m+1) C=mA = (3m + 1) \\\ B = 2(m + 1) \\\ C = m
Using the above-mentioned concept we need to find discriminant (D), we get :
D=[2(m+1)]24×(3m+1)×(m) =4(m2+1+2m)4(3m2+m)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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
Now , D=0D = 0
4(m2+1+2m)4(3m2+m)=0 (m2+1+2m)=(3m2+m) m2+1+2m=3m2+m 2m2m1=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
2m22m+m1=0 2m(m1)+1(m1)=0 (2m+1)(m1)=0 2m+1=0,m1=0 m=1,122{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 1,121,\dfrac{{ - 1}}{2}.
So, option A is the correct answer.

Note: For any quadratic equation Ax2+Bx+C=0A{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=B24ACD = {B^2} - 4AC . Now the three types of roots are as follows:
(i). D>0 , Distinct and real roots exist and the roots are α=B+D2A,β=BD2A\alpha = \dfrac{{ - B + \sqrt D }}{{2A}},\beta = \dfrac{{ - B - \sqrt D }}{{2A}}
(ii). D=0, Real and equal roots exist and the roots are α=β=B2A\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.