(C) Given that $3$ is a root of the equation $x^2+kx-24=0$. Substituting $x=3$ into the equation: $(3)^2 + k(3) - 24 = 0$ $9 + 3k - 24 = 0$ $3k - 15 =

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(C) Given that $3$ is a root of the equation $x^2+kx-24=0$. Substituting $x=3$ into the equation: $(3)^2 + k(3) - 24 = 0$ $9 + 3k - 24 = 0$ $3k - 15 = 0$ $3k = 15 \Rightarrow k = 5$. Now,we check the options by substituting $k=5$ and $x=3$: For option $C$: $x^2 - kx + 6 = 0$ Substituting $x=3$ and $k=5$: $(3)^2 - (5)(3) + 6 = 9 - 15 + 6 = 0$. Since the equation is satisfied,$3$ is a root of $x^2-kx+6=0$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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If $3$ is a root of $x^2+kx-24=0$,then it is also a root of which of the following equations?

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A
$x^2+5x+k=0$
B
$x^2+kx+24=0$
C
$x^2-kx+6=0$
D
$x^2-5x+k=0$

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Solution of quadratic equations and Nature of roots

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