Question

the value of K so that the equation 2x^2+kx-5=0 and x^2-3x-4 have one root in common is

Answer 1

=x^2-3x-4
=x^2-4x+x-4
=x(x-4) +1(x-4)
=(x-4) (x+1)
x= 4 and -1

2x^2+kx-5=0
x=4
2*4^2+4k-5=0
32-5+4k=0
4k = 27
k = 27/4
or
x = -1
2*1-k-5=0
-k -5+2=0
-k = 3
k = -3

Answer 2

• The eq. are 2$x^{2}$+kx-5=0 --------(1)

                       and $x^{2}$-3x-4=0 --------(2)

• We will find roots of eq. (2)

        $x^{2}$-3x-4=0

        $x^{2}$-4x+x-4=0

        (x+1)(x-4)=0

        x=-1,4

• Now, if the common root is -1

       then putting -1 in eq(1)

       2*$(-1)^{2}$+k(-1)-5=0

       2-5-k=0

       k=3

• if the common root is 4

      then putting 4 in eq(1)

      2*$4^{2}$+k(4)-5=0

      32+4k-5=0

       k=-27/4