Finding Conditions for Real Roots of Quadratic Equations

The question asks for the condition under which the quadratic equation \(x^2 + mx + 2 = 0\) will always have real roots. A quadratic equation in the standard form \(ax^2 + bx + c = 0\) has real roots if and only if its discriminant is non-negative. The discriminant, denoted by the symbol \(\Delta\), is calculated using the formula \(\Delta = b^2 - 4ac\).

Understanding the Discriminant and Real Roots

• The discriminant (\(\Delta\)) tells us about the nature of the roots of a quadratic equation.
• If \(\Delta > 0\), the equation has two distinct real roots.
• If \(\Delta = 0\), the equation has exactly one real root (or two equal real roots).
• If \(\Delta < 0\), the equation has two complex conjugate roots (not real roots).

For the quadratic equation to have real roots, the condition is \(\Delta \ge 0\).

Applying the Condition to the Given Equation

The given quadratic equation is \(x^2 + mx + 2 = 0\).

Comparing this with the standard form \(ax^2 + bx + c = 0\), we can identify the coefficients:

• \(a = 1\)
• \(b = m\)
• \(c = 2\)

Now, let's calculate the discriminant for this equation:

$\Delta = b^2 - 4ac$

Substitute the values of \(a\), \(b\), and \(c\):

$\Delta = m^2 - 4 \times 1 \times 2$

$\Delta = m^2 - 8$

For the equation to have real roots, the discriminant must be greater than or equal to zero:

$\Delta \ge 0$

So, we must have:

$m^2 - 8 \ge 0$

Adding 8 to both sides of the inequality, we get:

$m^2 \ge 8$

This is the condition on \(m\) for the quadratic equation \(x^2 + mx + 2 = 0\) to always have real roots. Let's compare this condition with the given options.

Comparing with Options

Let's look at the options provided:

1. $2\sqrt{3} \le m^2 < 8$
2. $\sqrt{3} \le m^3 < 4$
3. $m^2 \ge 8$
4. $m^2 \le \sqrt{3}$

The condition we derived is \(m^2 \ge 8\). This matches option 3.

Conclusion on Real Roots Condition

For the quadratic equation \(x^2 + mx + 2 = 0\) to have real roots, the discriminant must be greater than or equal to zero. Calculating the discriminant gives \(\Delta = m^2 - 8\). Setting \(\Delta \ge 0\) leads to the inequality \(m^2 - 8 \ge 0\), which simplifies to \(m^2 \ge 8\). Therefore, the quadratic equation will always have real roots if and only if \(m^2 \ge 8\).

Revision Table: Key Concepts for Real Roots

Additional Information: Quadratic Equations and Root Nature

A quadratic equation is a polynomial equation of the second degree. Its general form is \(ax^2 + bx + c = 0\), where \(x\) is the variable, and \(a\), \(b\), and \(c\) are coefficients with \(a \ne 0\).

The nature of the roots (whether they are real, complex, distinct, or equal) is entirely determined by the value of the discriminant, \(\Delta = b^2 - 4ac\).

• If \(\Delta\) is a perfect square and non-negative, the roots are real and rational.
• If \(\Delta\) is positive but not a perfect square, the roots are real and irrational.
• If \(\Delta\) is zero, the roots are real, rational, and equal.
• If \(\Delta\) is negative, the roots are complex conjugates and not real.

In this problem, we are only concerned with the condition for real roots, which means \(\Delta\) must be non-negative (\(\Delta \ge 0\)). The value of \(m\) affects the coefficient \(b\), and thus the value of the discriminant, determining the nature of the roots.