If $x = 2$ and $x = 3$ are roots of the equation $3{x^2} - 2mx - 2n = 0$ find the value of $m$ and $n$ .
Answer
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Hint: As we know that the roots of quadratic equation satisfy the given equation so $x = 2$ and $x = 3$ are roots of the equation $3{x^2} - 2mx - 2n = 0$ now put the value of $x = 2$ and $x = 3$ two equation will form in m and n Solve these equation and get the value of m and n.
Complete step-by-step answer:
As in the question the roots of the equation is given and the equation is $3{x^2} - 2mx - 2n = 0$ and its root is $x = 2$ and $x = 3$ .
Hence we know that the root of the equation is satisfy the given equation ,
So as it is given that $x = 2$ of equation $3{x^2} - 2mx - 2n = 0$ , Now put the value of $x = 2$ in this ,
we get ,
$3.{(2)^2} - 2m(2) - 2n = 0$
Divide by $2$ in whole equation we get ,
$\Rightarrow$ $6 - 2m - n = 0$ ,
$\Rightarrow$ $2m + n = 6$ ..........(i)
Now as it is given that $x = 3$ of equation $3{x^2} - 2mx - 2n = 0$ , Now put the value of $x = 3$ in this ,
we get ,
$\Rightarrow$ $3.{(3)^2} - 2m(3) - 2n = 0$
On solving and rearranging we get ,
$\Rightarrow$ $6m + 2n = 27$ .........(ii)
Now we have to find the value of $m$ and $n$ , from equation $2m + n = 6$ ..........(i)
$\Rightarrow$ $6m + 2n = 27$ .........(ii)
On multiplying by $2$ in equation .....(i) we get $4m + 2n = 12$
Now subtract equation (i) from (ii) we get ,
we get , $4m - 6m = 12 - 27$
$\Rightarrow$ $ - 2m = - 15$
Hence $m = \dfrac{{15}}{2}$
Now put the value of m in the equation .....(i) for the value of n
$4\left( {\dfrac{{15}}{2}} \right) + 2n = 12$
$\Rightarrow$ $2n = 12 - 30$
On solving we get , $n = - 9$
So from the above solution we get $m = \dfrac{{15}}{2}$ and $n = - 9$ .
Note: This question we can also solve by the alternative method that is if the quadratic equation is $a{x^2} + bx + c = 0$ then , Sum of roots is equal to $\dfrac{{ - b}}{a}$ and product of roots is $\dfrac{c}{a}$ Roots of equation is given that is $2$ and $3$ so from sum of roots ,$5 = \dfrac{{ - ( - 2m)}}{3}$ and from product of roots $6 = \dfrac{{ - 2n}}{3}$ , From this we can find the value of m and n .
Complete step-by-step answer:
As in the question the roots of the equation is given and the equation is $3{x^2} - 2mx - 2n = 0$ and its root is $x = 2$ and $x = 3$ .
Hence we know that the root of the equation is satisfy the given equation ,
So as it is given that $x = 2$ of equation $3{x^2} - 2mx - 2n = 0$ , Now put the value of $x = 2$ in this ,
we get ,
$3.{(2)^2} - 2m(2) - 2n = 0$
Divide by $2$ in whole equation we get ,
$\Rightarrow$ $6 - 2m - n = 0$ ,
$\Rightarrow$ $2m + n = 6$ ..........(i)
Now as it is given that $x = 3$ of equation $3{x^2} - 2mx - 2n = 0$ , Now put the value of $x = 3$ in this ,
we get ,
$\Rightarrow$ $3.{(3)^2} - 2m(3) - 2n = 0$
On solving and rearranging we get ,
$\Rightarrow$ $6m + 2n = 27$ .........(ii)
Now we have to find the value of $m$ and $n$ , from equation $2m + n = 6$ ..........(i)
$\Rightarrow$ $6m + 2n = 27$ .........(ii)
On multiplying by $2$ in equation .....(i) we get $4m + 2n = 12$
Now subtract equation (i) from (ii) we get ,
we get , $4m - 6m = 12 - 27$
$\Rightarrow$ $ - 2m = - 15$
Hence $m = \dfrac{{15}}{2}$
Now put the value of m in the equation .....(i) for the value of n
$4\left( {\dfrac{{15}}{2}} \right) + 2n = 12$
$\Rightarrow$ $2n = 12 - 30$
On solving we get , $n = - 9$
So from the above solution we get $m = \dfrac{{15}}{2}$ and $n = - 9$ .
Note: This question we can also solve by the alternative method that is if the quadratic equation is $a{x^2} + bx + c = 0$ then , Sum of roots is equal to $\dfrac{{ - b}}{a}$ and product of roots is $\dfrac{c}{a}$ Roots of equation is given that is $2$ and $3$ so from sum of roots ,$5 = \dfrac{{ - ( - 2m)}}{3}$ and from product of roots $6 = \dfrac{{ - 2n}}{3}$ , From this we can find the value of m and n .
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