Let f(x) =x^2+Kx and g(x) = x+K. The graphs of f and g intersect at two distinct points. Find the value of K?

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Answer 1 · 2017-04-24T07:31:56

For graphs $f (x)$ $f(x)$ and $g (x)$ $g(x)$ to intersect at two distinct points, we must have $k \neq - 1$ $k!=-1$

As $f (x) = x^{2} + k x$ $f(x)=x^2+kx$ and $g (x) = x + k$ $g(x)=x+k$

and they will intersect where $f (x) = g (x)$ $f(x)=g(x)$

or $x^{2} + k x = x + k$ $x^2+kx=x+k$

or $x^{2} + k x - x - k = 0$ $x^2+kx-x-k=0$

As this has two distinct solutions,

the discriminant of quadratic equation must be greater than $0$ $0$ i.e.

${(k - 1)}^{2} - 4 \times (- k) > 0$ $(k-1)^2-4xx(-k)>0$

or ${(k - 1)}^{2} + 4 k > 0$ $(k-1)^2+4k>0$

or ${(k + 1)}^{2} > 0$ $(k+1)^2>0$

As ${(k + 1)}^{2}$ $(k+1)^2$ is always greater than $0$ $0$ except when $k = - 1$ $k=-1$

Hence, for graphs $f (x)$ $f(x)$ and $g (x)$ $g(x)$ to intersect at two distinct points, we must have $k \neq - 1$ $k!=-1$