Question

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

• A. 7
• B. 8
• C. 9
• D. 13

Answer 1

Answer: (C)

9

Answer 2

The correct answer is C:9
To solve this problem,we need to analyze the conditions for both equations to have no real roots and two distinct real roots,respectively.Let's break it down step by step.
Equation 1: $2x^2 + kx + 5 = 0$
For this equation to have no real roots,the discriminant $(b^2 - 4ac)$ must be negative,where the coefficients are from the quadratic equation $ax^2+bx+c=0$
In this case, a=2,b=k, and c=5. So,the discriminant is:
$D1 = k^2 - 4 \times 2 \times 5 = k^2 - 40.$
For no real roots, D1<0:
$k^2 - 40 < 0$
$k^2 < 40$
$-\sqrt{40} < k < \sqrt{40}$
$-2\sqrt{10} < k < 2\sqrt{10}$
Equation 2: $x^2 + (k - 5)x + 1 = 0$
For this equation to have two distinct real roots,the discriminant must be positive.
In this case, a=1, b=k-5, and c=1.So,the discriminant is:
$D2 = (k - 5)^2 - 4 \times 1 \times 1 = k^2 - 10k + 21$.
For two distinct real roots, D2>0:
$k^2 - 10k + 21 > 0$
(k-3)(k-7)>0
This quadratic inequality holds when k<3 or k>7.
Now,let's combine the conditions we found for both equations:
$-2\sqrt{10} < k < 2\sqrt{10}$ (for no real roots in the first equation)
k<3 or k>7 (for two distinct real roots in the second equation)
To find the possible values of k that satisfy both conditions,we need to find the intersection of the intervals:
$-2\sqrt{10} < k < 2\sqrt{10}$k < 3 or k > 7
Visualizing this on a number line:
-∞ ---- ($-2\sqrt{10}$) -------- 3 -------- 7 -------- ($2\sqrt{10}$) ---- +∞
The values that satisfy both conditions are within the range of $-2\sqrt{10}\space to\space 2\sqrt{10}$ and also either less than 3 or greater than 7.
Thus, the possible values of k are within the following ranges:
From $-2\sqrt{10}$ to 3 (excluding 3)
From 7 to $2\sqrt{10}$ (excluding 7)
Calculating these ranges:
$-2\sqrt{10}$≈-8.94, so the range is approximately -8.94<k<3.
$2\sqrt{10}$≈8.94, so the range is approximately 7<k<8.94.
Therefore,there are 9 possible integer values of k that satisfy both conditions: -8,-7,-6,-5,-4,-3,-2,-1, and 0.
Hence,the answer is indeed 9 possible values of k.