Understanding Real Roots of a Quadratic Equation

A quadratic equation in the standard form \(Ax^2 + Bx + C = 0\) has real roots if and only if its discriminant, denoted by \(\Delta\) or \(D\), is greater than or equal to zero. The discriminant is calculated using the formula: \(\Delta = B^2 - 4AC\).

Analyzing the Given Equation

The given quadratic equation is \(x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0\). Comparing this equation with the standard form \(Ax^2 + Bx + C = 0\), we can identify the coefficients:

• \(A = 1\)
• \(B = -(1 - 2a^2)\)
• \(C = (1 - 2a^2)\)

Calculating the Discriminant

Now, we calculate the discriminant \(\Delta\) using the identified coefficients:

\(\Delta = B^2 - 4AC\)

Substitute the values of A, B, and C:

\(\Delta = (-(1 - 2a^2))^2 - 4(1)(1 - 2a^2)\)

Simplifying the expression:

\(\Delta = (1 - 2a^2)^2 - 4(1 - 2a^2)\)

Setting the Condition for Real Roots

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

\(\Delta \ge 0\)

So, we must have:

\((1 - 2a^2)^2 - 4(1 - 2a^2) \ge 0\)

Solving the Inequality for 'a'

Let's solve this inequality to find the condition on \(a^2\). We can factor out the common term \((1 - 2a^2)\):

\((1 - 2a^2)[(1 - 2a^2) - 4] \ge 0\)

\((1 - 2a^2)(-3 - 2a^2) \ge 0\)

We can rewrite \((-3 - 2a^2)\) as \(-(3 + 2a^2)\). So the inequality becomes:

\((1 - 2a^2)[-(3 + 2a^2)] \ge 0\)

\(-(1 - 2a^2)(3 + 2a^2) \ge 0\)

Multiply both sides by -1 and reverse the inequality sign:

\((1 - 2a^2)(3 + 2a^2) \le 0\)

Now, let's analyze the factors:

• The term \((3 + 2a^2)\) involves \(a^2\), which is always non-negative (\(a^2 \ge 0\)) for any real number \(a\). Therefore, \(2a^2 \ge 0\), and \(3 + 2a^2 \ge 3\). This means the factor \((3 + 2a^2)\) is always positive.
• For the product \((1 - 2a^2)(3 + 2a^2)\) to be less than or equal to zero, the other factor \((1 - 2a^2)\) must be less than or equal to zero, since \((3 + 2a^2)\) is always positive.

So, we must have:

\(1 - 2a^2 \le 0\)

Add \(2a^2\) to both sides:

\(1 \le 2a^2\)

Divide both sides by 2:

\(\frac{1}{2} \le a^2\)

This inequality can also be written as:

\(a^2 \ge \frac{1}{2}\)

Conclusion

The condition under which the given quadratic equation \(x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0\) has real roots is \(a^2 \ge \frac{1}{2}\).