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Question

If $$3$$ is a root of $$x^2+kx-24=0$$ then $$3$$ is also root of the equation


A
x2+5x+k=0
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B
x2+kx+24=0
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C
x2kx+6=0
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D
x25x+k=0
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Solution

The correct option is C $$x^{2}-kx+6=0$$
$$x^{2}+kx-24=0$$...(i)
It is given that $$x=3$$ satisfies the equation.
Hence
$$9+3k-24=0$$
$$3k=24-9$$
$$3k=15$$
$$k=5$$
Thus the equation is
$$x^{2}+5x-24=0$$
$$(x+8)(x-3)=0$$
Hence the roots are $$x=-8,3$$.
Now $$k=5$$.
Hence an equation of the form
$$x^{2}-kx+d=0$$ is
$$x^{2}-5x+d=0$$.
It is given that $$x=3$$ satisfies that equation.
Hence
$$9-15+d=0$$
$$d=6$$.
Thus one possible equation is
$$x^{2}-5x+6=0$$
Or
$$x^{2}-kx+6=0$$

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