Question

If 3 is a root of ${{x}^{2}}+kx-24=0$ then 3 is also the root of the equation
( a ) ${{x}^{2}}+5x+k=0$
( b ) ${{x}^{2}}+kx+24=0$
( c ) ${{x}^{2}}-kx+6=0$
( d ) ${{x}^{2}}-5x+k=0$

Answer 1

Hint: In this question, we are given a quadratic equation with one of its roots and we have to find the other quadratic equation with the same root. By putting the root in the equation, we find the value of k and hence we find out the real roots of the quadratic equation. Then we take another equation which is in the form of d and finding out the value of d, we get another quadratic equation.

Formula Used: Standard quadratic equation = $a{{x}^{2}}+bx+c=0$

Complete step by step Solution:
Given equation is ${{x}^{2}}+kx-24=0$
Given 3 is the one root of the above equation
So ${{(3)}^{2}}+3k-24=0$
That is 3k – 15 = 0
And k = 5
Thus, the quadratic equation is $a{{x}^{2}}+bx+c=0$
And ${{x}^{2}}+8x-3x-24=0$
By solving it, we get (x + 8 )( x – 3 ) = 0
Hence the roots are x = -8, 3
As the value of k = 5
Then the equation is in the form${{x}^{2}}-kx+d=0$
That is ${{x}^{2}}-5x+d=0$
As we find it, x = 3 satisfies the equation
Hence ${{(3)}^{2}}-5(3)+d=0$
And the value of d = 6
Thus one possible equation is ${{x}^{2}}-5x+6=0$ OR
                                                      ${{x}^{2}}-kx+6=0$

Therefore, the correct option is (c).

Note: In these types of questions, students made mistakes in finding out the quadratic equation. They only find the value of k and put the value of k and think that it is the answer but in these types of questions, we then form the equation in the form of d, and finding the value of d we find the quadratic equation.