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

The Correct Option is C

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.

Answer 2

The given equation,
\(2x^2 + kx + 5 = 0\)
The equation has no real roots,
\(⇒\) \(D<0\)
\(⇒(K-\sqrt 40)(K+\sqrt 40)<0\)
\(⇒K∈(-\sqrt {40}, \sqrt {40})\)

\(x^2 + (k - 5)x + 1 = 0\)
It has two distinct real roots,
\(⇒D>0\)
\(⇒(k - 5)^2 — 4 > 0\)
\(⇒k^2 — 10k + 21 > 0\)
\(⇒(k - 3) (k - 7) > 0\)
\(⇒k ∈ (-∞ , 3)\) U \((7, ∞ )\)
Thus, the possibe value of \(k = -6, -5, -4, -3, -2, -1, 0, 1, 2\) i.e. \(9\) integers

So, the correct option is (C): \(9\)