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Question

Find the value of $$k$$ for which $$x=3$$ is a solution of the quadratic equation, $$\left(k+2\right){x}^{2}-kx+6=0$$.
Thus find the other root of the equation.


Solution

Given : $$\left( k+2 \right) { x }^{ 2 }-kx+6=0$$
Putting $$x=3$$
            $$\left( k+2 \right) \times 9-k\times 3+6=0$$
            $$9k+18-3k+6=0$$
            $$6k=-24$$
            $$k=-4$$
Putting $$k=-4$$ in given equation
            $$-2{ x }^{ 2 }+4x+6=0$$
            $${ x }^{ 2 }-2x-3=0$$
            $${ x }^{ 2 }-3x+x-3=0$$
            $$x\left( x-3 \right) +1\left( x-3 \right) =0$$
            $$\left( x+1 \right) \left( x-3 \right) =0$$
            $$x+1=0$$ or $$x-3=0$$
            $$x=-1$$ or $$x=3$$
Other root of the equation, $$x=-1$$

Mathematics

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