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: \[ D_1 = k^2 - 4(2)(5) = k^2 - 40 < 0 \Rightarrow k^2 < 40 \Rightarrow -\sqrt{40} < k < \sqrt{40} \Rightarrow -2\sqrt{10} < k < 2\sqrt{10} \] Since \( \sqrt{10} \approx 3.16 \), we have: \[ -6.32 < k < 6.32 \Rightarrow k \in (-6.32, 6.32) \]

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: \[ D_2 = (k - 5)^2 - 4(1)(1) = k^2 - 10k + 21 > 0 \] Factor the inequality: \[ k^2 - 10k + 21 > 0 \Rightarrow (k - 3)(k - 7) > 0 \Rightarrow k < 3 \text{ or } k > 7 \]

Step 3: Combine Both Conditions

From Step 1: \( k \in (-\sqrt{40}, \sqrt{40}) \approx (-6.32, 6.32) \) From Step 2: \( k < 3 \text{ or } k > 7 \) Intersection:

• For \( k < 3 \) and \( k \in (-6.32, 6.32) \): we get \( k \in (-6.32, 3) \)
• For \( k > 7 \): does not overlap with \( (-6.32, 6.32) \), so it's excluded

 

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

Valid integer values: \( -6, -5, -4, -3, -2, -1, 0, 1, 2 \) Total = \( \boxed{9} \)

Final Answer:

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

Answer 2

We are given two quadratic equations:

1. \( 2x^2 + kx + 5 = 0 \) has no real roots
2. \( x^2 + (k - 5)x + 1 = 0 \) has two distinct real roots

Step 1: No Real Roots for the First Equation

Use the discriminant condition \( D < 0 \) for no real roots: \[ D = k^2 - 4(2)(5) = k^2 - 40 < 0 \Rightarrow k \in (-\sqrt{40}, \sqrt{40}) \Rightarrow k \in (-2\sqrt{10}, 2\sqrt{10}) \] Approximating: \( \sqrt{10} \approx 3.16 \Rightarrow 2\sqrt{10} \approx 6.32 \) \[ \Rightarrow k \in (-6.32, 6.32) \]

Step 2: Two Distinct Real Roots for the Second Equation

Use the discriminant condition \( D > 0 \): \[ D = (k - 5)^2 - 4 > 0 \Rightarrow k^2 - 10k + 21 > 0 \Rightarrow (k - 3)(k - 7) > 0 \Rightarrow k \in (-\infty, 3) \cup (7, \infty) \]

Step 3: Combine Both Conditions

Intersection of:

• \( k \in (-6.32, 6.32) \)
• \( k \in (-\infty, 3) \cup (7, \infty) \)

Gives:

• \( k \in (-6.32, 3) \) — because \( (7, \infty) \) lies outside \( (-6.32, 6.32) \)

 

Step 4: Count Integer Values in \( (-6.32, 3) \)

Valid integers are: \[ -6, -5, -4, -3, -2, -1, 0, 1, 2 \] That’s a total of: \[ \boxed{9} \text{ integer values} \]

Final Answer:

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