Question 2

# 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

Solution

$$2x^{2}+kx+5=0$$ 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.