Let the given quadratic equation be

\[x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0\]

Let \(\alpha\) and \(\beta\) be the roots of this equation. For a general quadratic equation \(Ax^2 + Bx + C = 0\), Vieta's formulas give the sum and product of the roots:

• Sum of roots: \(\alpha + \beta = -\frac{B}{A}\)
• Product of roots: \(\alpha \beta = \frac{C}{A}\)

In our given equation, we have \(A=1\), \(B=-(1 - 2a^2)\), and \(C=(1 - 2a^2)\). Applying Vieta's formulas:

• Sum of roots: \(\alpha + \beta = -\frac{-(1 - 2a^2)}{1} = 1 - 2a^2\)
• Product of roots: \(\alpha \beta = \frac{1 - 2a^2}{1} = 1 - 2a^2\)

We are given the condition \(\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}} < 1\). First, let's rewrite the left side of the inequality in terms of the sum and product of the roots:

\[\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}} = \frac{{{\rm{\beta }}^2} + {{\rm{\alpha }}^2}}{{{{\rm{\alpha }}^2}{{\rm{\beta }}^2}}}\]

We know that \({{\rm{\alpha }}^2} + {{\rm{\beta }}^2} = {{({\rm{\alpha }} + {\rm{\beta }})}^2} - 2{\rm{\alpha }}{\rm{\beta }}\). Also, \({{\rm{\alpha }}^2}{{\rm{\beta }}^2} = {{({\rm{\alpha }}{\rm{\beta }})}^2}\).

So, the inequality becomes:

\[\frac{{{({\rm{\alpha }} + {\rm{\beta }})}^2} - 2{\rm{\alpha }}{\rm{\beta }}}{{{{({\rm{\alpha }}{\rm{\beta }})}^2}}} < 1\]

Now, substitute the expressions for \({\rm{\alpha }} + {\rm{\beta }}\) and \({\rm{\alpha }}{\rm{\beta }}\) in terms of \({\rm{a}}\):

\[\frac{{{(1 - 2{{\rm{a}}^2})}^2} - 2(1 - 2{{\rm{a}}^2})}{{{{(1 - 2{{\rm{a}}^2})}^2}}} < 1\]

Let \(y = 1 - 2{{\rm{a}}^2}\) for simplicity. The inequality is:

\[\frac{{{y}^2} - 2y}{{{y}^2}} < 1\]

Note that for the terms \(\frac{1}{{{\rm{\alpha }}^2}}}\) and \(\frac{1}{{{{\rm{\beta }}^2}}}\) to be defined, the roots \(\alpha\) and \(\beta\) must be non-zero. This implies that the product of roots \({\rm{\alpha }}{\rm{\beta }}\) must be non-zero. Thus, \(1 - 2{{\rm{a}}^2} \ne 0\), which means \(y \ne 0\).

Since \(y \ne 0\), we can simplify the inequality:

\[\frac{{{y}^2} - 2y}{{{y}^2}} - 1 < 0\]

\[\frac{{{y}^2} - 2y - {{y}^2}}{{{y}^2}} < 0\]

\[\frac{-2y}{{{y}^2}} < 0\]

\[\frac{-2}{y} < 0\]

For \(\frac{-2}{y} < 0\), since the numerator \(-2\) is negative, the denominator \(y\) must be positive. Therefore, we need \(y > 0\).

Substitute back \(y = 1 - 2{{\rm{a}}^2}\):

\[1 - 2{{\rm{a}}^2} > 0\]

\[1 > 2{{\rm{a}}^2}\]

\[{{\rm{a}}^2} < \frac{1}{2}\]

This condition \({\rm{a}}^2 < \frac{1}{2}\) also ensures that \(1 - 2{{\rm{a}}^2} > 0\), which means \(y \ne 0\), satisfying the requirement that the roots are non-zero.

We should also consider the nature of the roots. The discriminant of the quadratic equation is \(\Delta = B^2 - 4AC = {(-(1 - 2{{\rm{a}}^2}))^2} - 4(1)(1 - 2{{\rm{a}}^2}) = {(1 - 2{{\rm{a}}^2})}^2 - 4(1 - 2{{\rm{a}}^2})\). \(\Delta = (1 - 2{{\rm{a}}^2})[(1 - 2{{\rm{a}}^2}) - 4] = (1 - 2{{\rm{a}}^2})(-3 - 2{{\rm{a}}^2})\).

For \({\rm{a}}^2 < \frac{1}{2}\), we have \(1 - 2{{\rm{a}}^2} > 0\). Since \({\rm{a}}^2 \ge 0\), \(-3 - 2{{\rm{a}}^2}\) is always negative. Thus, \(\Delta = (1 - 2{{\rm{a}}^2})(-3 - 2{{\rm{a}}^2})\) will be negative when \({\rm{a}}^2 < \frac{1}{2}\). A negative discriminant means the roots are complex conjugates. If the roots are complex conjugates \(p \pm iq\) with \(q \ne 0\), their product is \(p^2 + q^2\), which is always positive. This matches \(1 - 2{{\rm{a}}^2} > 0\). Also, if the roots are complex, they cannot be zero, so \(\alpha \ne 0\) and \(\beta \ne 0\) is guaranteed if \(1 - 2a^2 \ne 0\). If \(1 - 2a^2 > 0\), then the roots are non-zero complex numbers, and our derivation holds.

If \({\rm{a}}^2 > \frac{1}{2}\), then \(1 - 2{{\rm{a}}^2} < 0\). In this case, \(-3 - 2{{\rm{a}}^2}\) is negative, so \(\Delta = (1 - 2{{\rm{a}}^2})(-3 - 2{{\rm{a}}^2})\) will be positive, meaning real and distinct roots. The product of roots \(1 - 2{{\rm{a}}^2}\) is negative. Our inequality \(\frac{-2}{y} < 0\) requires \(y > 0\), so \(1 - 2{{\rm{a}}^2} > 0\). This contradicts \(1 - 2{{\rm{a}}^2} < 0\) for \({\rm{a}}^2 > \frac{1}{2}\).

If \({\rm{a}}^2 = \frac{1}{2}\), then \(1 - 2{{\rm{a}}^2} = 0\). The equation becomes \(x^2 = 0\), with roots \(\alpha = \beta = 0\). The expression \(\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}}\) is undefined.

Thus, the condition for the inequality \(\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}} < 1\) is \({\rm{a}}^2 < \frac{1}{2}\).

Revision Table: Roots of Quadratic Equations and Vieta's Formulas

Additional Information: Manipulating Root Expressions

Expressions involving powers of roots like \(\alpha^2 + \beta^2\) or \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\) can often be simplified using symmetric polynomial identities and Vieta's formulas.

• \({{\rm{\alpha }}^2} + {{\rm{\beta }}^2} = {{({\rm{\alpha }} + {\rm{\beta }})}^2} - 2{\rm{\alpha }}{\rm{\beta }}\)
• \({{\rm{\alpha }}^3} + {{\rm{\beta }}^3} = {{({\rm{\alpha }} + {\rm{\beta }})}^3} - 3{\rm{\alpha }}{\rm{\beta }}({\rm{\alpha }} + {\rm{\beta }})\)
• \(\frac{1}{{\rm{\alpha }}} + \frac{1}{{\rm{\beta }}} = \frac{{\rm{\beta }} + {\rm{\alpha }}}{{\rm{\alpha }}{\rm{\beta }}}\)
• \(\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}} = \frac{{{\rm{\beta }}^2} + {{\rm{\alpha }}^2}}{{{{({\rm{\alpha }}{\rm{\beta }})}^2}}} = \frac{{{({\rm{\alpha }} + {\rm{\beta }})}^2} - 2{\rm{\alpha }}{\rm{\beta }}}{{{{({\rm{\alpha }}{\rm{\beta }})}^2}}}\)

These identities are useful for solving problems involving roots without actually finding the values of the roots themselves, relying only on the coefficients of the quadratic equation.

The condition that \(\frac{1}{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}}\) exists requires \(\alpha \ne 0\) and \(\beta \ne 0\). This implies the constant term of the quadratic equation must be non-zero, i.e., \(C \ne 0\). In this problem, \(C = 1 - 2a^2\), so we must have \(1 - 2a^2 \ne 0\), meaning \(a^2 \ne 1/2\).

The final answer is the condition derived from the inequality: \({\rm{a}}^2 < \frac{1}{2}\).