
If the product of roots of the equation $m{{x}^{2}}+6x+(2m-1)=0$ is -1, then the value
of m is
( a ) $\dfrac{1}{3}$
( b ) 1
( c ) 3
( d ) – 1
Answer
161.1k+ views
Hint: Here we are given the equation and the product of roots. We have to notice the value of m. we know the product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ . By putting the values in the equation and equating it with -1, we are able to find out the value of m.
Formula Used: Product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
Complete step by step Solution:
We have given the equation $m{{x}^{2}}+6x+(2m-1)=0$
We have to notice the value of m.
Given the product of roots of the above equation is – 1.
As we know according to the formula,
Product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
That is $\dfrac{2m-1}{m}=-1$
Solving the above equation, we get
2m-1 = -m
Then 3m = 1
That is $m=\dfrac{1}{3}$
Hence, the value of $m=\dfrac{1}{3}$
Therefore, the correct option is (a).
Note: We know in a quadratic equation, there are two roots which are the sum of roots and the product of roots. We know sum of roots = $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and the product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ . By putting the values in the equations, we get the roots of the quadratic equation. We must try to solve the equation by the easiest way and for that prior inspection will be needed rather than to start solving questions without thinking of the method.
Formula Used: Product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
Complete step by step Solution:
We have given the equation $m{{x}^{2}}+6x+(2m-1)=0$
We have to notice the value of m.
Given the product of roots of the above equation is – 1.
As we know according to the formula,
Product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
That is $\dfrac{2m-1}{m}=-1$
Solving the above equation, we get
2m-1 = -m
Then 3m = 1
That is $m=\dfrac{1}{3}$
Hence, the value of $m=\dfrac{1}{3}$
Therefore, the correct option is (a).
Note: We know in a quadratic equation, there are two roots which are the sum of roots and the product of roots. We know sum of roots = $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and the product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ . By putting the values in the equations, we get the roots of the quadratic equation. We must try to solve the equation by the easiest way and for that prior inspection will be needed rather than to start solving questions without thinking of the method.
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