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
B
x2+kx+24=0
C
x2kx+6=0
D
x25x+k=0

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 $$x^{2}-kx+6=0$$Maths

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