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

Step 1: Condition for No Real Roots in Equation 1

For the equation \( 2x^2 + kx + 5 = 0 \), the discriminant must be negative. The discriminant is calculated as \( D_1 = k^2 - 4(2)(5) = k^2 - 40 \). For no real roots, \( k^2 - 40<0 \), which simplifies to \( k^2<40 \). This implies \( -\sqrt{40}<k<\sqrt{40} \), which is equivalent to \( -2\sqrt{10}<k<2\sqrt{10} \). Since \( \sqrt{10} \approx 3.16 \), the range for \( k \) is approximately \( -6.32<k<6.32 \), or \( k \in (-6.32, 6.32) \).

Step 2: Condition for Two Distinct Real Roots in Equation 2

For the equation \( x^2 + (k - 5)x + 1 = 0 \), the discriminant must be positive to have two distinct real roots. The discriminant is \( D_2 = (k - 5)^2 - 4(1)(1) = k^2 - 10k + 21 \). For two distinct real roots, \( k^2 - 10k + 21>0 \). Factoring the quadratic, we get \( (k - 3)(k - 7)>0 \). This inequality holds when \( k<3 \) or \( k>7 \).

Step 3: Combine Both Conditions

The conditions derived are: From Step 1: \( k \in (-\sqrt{40}, \sqrt{40}) \approx (-6.32, 6.32) \). From Step 2: \( k<3 \) or \( k>7 \). We need to find the intersection of these conditions.

• For \( k<3 \) and \( k \in (-6.32, 6.32) \), the intersection is \( k \in (-6.32, 3) \).
• For \( k>7 \), this range does not overlap with \( (-6.32, 6.32) \), so there are no values of \( k \) that satisfy both conditions in this case.

Step 4: Count Integer Values in the Range \( -6.32 < k < 3 \)

The integers within the range \( -6.32<k<3 \) are: \( -6, -5, -4, -3, -2, -1, 0, 1, 2 \). There are a total of \( 9 \) integer values.

Final Answer:

\[ \boxed{9} \quad \text{(Correct Option: C)} \]