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 have equal root?
A). 1,2−1
B). 2 or 4
C). 4
D). 3
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
Complete Step by step solution:
We know for any quadratic equation Ax2+Bx+C=0, equal roots are possible only when discriminant(D) = 0. Which can be found by the formula D=B2−4AC.
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=[2(m+1)]2−4×(3m+1)×(m) =4(m2+1+2m)−4(3m2+m)
Here, we used the formula (a+b)2=a2+b2+2ab
Now , D=0
⇒4(m2+1+2m)−4(3m2+m)=0 ⇒(m2+1+2m)=(3m2+m) ⇒m2+1+2m=3m2+m ⇒2m2−m−1=0
Splitting the middle term, we get
2m2−2m+m−1=0 ⇒2m(m−1)+1(m−1)=0 ⇒(2m+1)(m−1)=0 ⇒2m+1=0,m−1=0 ⇒m=1,2−1
Therefore, the values of m are 1,2−1.
So, option A is the correct answer.
Note: For any quadratic equation Ax2+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=B2−4AC . Now the three types of roots are as follows:
(i). D>0 , Distinct and real roots exist and the roots are α=2A−B+D,β=2A−B−D
(ii). D=0, Real and equal roots exist and the roots are α=β=2A−B
(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.