Question

11 The roots of the equation \( x^{2}+k x+3=0 \) are real if the value of \( k \) is 1) 0 2) 2 3) 3 4) 4

Ask by Hilton Nichols. in the United States
Mar 17,2025

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The roots are real when \( k = 4 \).

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The Deep Dive

To determine when the roots of the quadratic equation \( x^{2}+kx+3=0 \) are real, we can use the discriminant condition. The discriminant \( D \) is given by the formula \( D = b^2 - 4ac \). For the roots to be real, \( D \) must be non-negative (i.e., \( D \geq 0 \)). In this equation, \( a = 1 \), \( b = k \), and \( c = 3 \). Therefore, the discriminant becomes: \[ D = k^2 - 4 \cdot 1 \cdot 3 = k^2 - 12 \] Now, we need to solve the inequality: \[ k^2 - 12 \geq 0 \] This can be rewritten as: \[ k^2 \geq 12 \] Taking square roots: \[ |k| \geq \sqrt{12} = 2\sqrt{3} \approx 3.46 \] Thus, \( k \) must be either \( k \leq -2\sqrt{3} \) or \( k \geq 2\sqrt{3} \). Now, let’s evaluate the given options: 1) \( k = 0 \) → Not valid. 2) \( k = 2 \) → Not valid. 3) \( k = 3 \) → Valid since \( 3 > 2\sqrt{3} \). 4) \( k = 4 \) → Also valid since \( 4 > 2\sqrt{3} \). So, the valid options for \( k \) that provide real roots are: **3 and 4.**

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