1.

Find the value of `k`, so that the equation `2x^2+kx-5=0` and `x^2-3x-4=0` may have one root in common.

Answer» `x^2-3x-4=0`
`=>x^2-4x+x-4=0`
`=>x(x-4)+1(x-4) = 0`
`=>(x-4)(x+1) = 0`
`=> x=4 and x=-1`
Now, as `2x^2+kx-5=0`, have one common root with the previous equation.So, values of `x` will satisfy this equation.
When `x = 4`
`2(4)^2+4k -5 = 0`
`32+4k = 5=>k =-27/4`
When `x = -1`
`2(-1)^2+(-1)k -5 = 0`
`2-k=5=> k = -3`
So, for `k=-27/4 and k = -3`, both equations have one root in common.


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