Question

If 2 and 3 are zeroes of polynomial 3x 2 - 2kx +2m find the value of k and m.

Answer 1

Please find the solution in the image

Answer 2

The value of (k,m) is (0,-6) .

Let $f(x)=3 x^{2}-2 k x+2 m \rightarrow(1)$

Given equation as 2 and 3 zeros of polynomial.  

That is f (2) = 0 and f (3) = 0

Substituting x = 2 in equation (1)

$f(x)=3 x^{2}-2 k x+2 m$

$f(2)=3(2)^{2}-2 k(2)+2 m$

$f(2)=3(4)-4 k+2 m$

$-4 k+2 m+12=0$

$4 k-2 m-12=0 \rightarrow(2)$

Substituting x = 3 in equation (2)

$\begin{array}{l}{f(x)=3 x^{2}-2 k x+2 m} \\ {f(3)=3(3)^{2}-2 k(3)+2 m}\end{array}$

$\begin{array}{l}{=3 \times 9-6 k+2 m} \\ {=12-6 k+2 m} \\ {6 k-2 m-12=0 \rightarrow(3)}\end{array}$

Solving equation (2) and (3)

Equation (2) becomes

$\begin{array}{l}{4 k-2 m=12} \\ {4 k=12+2 m} \\ {k=\frac{12+2 m}{4}=\frac{6+m}{2} \rightarrow(4)}\end{array}$

Substituting k = 6+m/2 in equation (3)

$\begin{array}{l}{6 k-2 m-12=0} \\ {\frac{6(6+m)}{2}-2 m-12=0} \\ {\frac{36+6 m}{2}-2 m-12=0}\end{array}$

$\begin{array}{l}{18+3 m-2 m-12=0} \\ {18+m-12=0} \\ {6+m=0} \\ {m=-6}\end{array}$

Substituting m = -6  in equation (4)

$\begin{array}{l}{k=\frac{6+m}{2}=\frac{6-6}{2}=0} \\ {k=0}\end{array}$

Therefore, the value of (k,m) is (0,-6) .