Suppose k is any integer such that the equation $$2x^{2}+kx+5=0$$ has no real roots and the equation $$x^{2}+(k-5)x+1=0$$ has two distinct real roots for x. Then, the number of possible values of k is
$$2x^{2}+kx+5=0$$ has no real roots so D<0
$$k^2-40\ <0$$
$$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$$
$$k\in\left(-\sqrt{40},\sqrt{40}\right)$$
$$x^{2}+(k-5)x+1=0$$ has two distinct real roots so D>0
$$\left(k-5\right)^2-4>0$$
$$k^2-10k+21>0$$
$$\left(k-3\right)\left(k-7\right)>0$$
$$k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$$
Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2
In 9 total 9 integer values of k are possible.
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