Min Value of $a^2+b^2$ in Quadratic Equation

[anemone]

Determine the minimum value of $a^2+b^2$ when $(a,\,b)$ traverses all the pairs of real numbers for which the equation $x^4+ax^3+bx^2+ax+1=0$ has at least one real root.

 

[SquareOne1]

Here's my attempt at a solution.

Spoiler: My Solution

Since \( x^{4} +ax^{3}+bx^{2}+ax+1 = 0 \), then we can assume that \( x \neq 0 \).

Dividing by \( x^{2} \) yields:
\( x^{2} + ax + \frac{a}{x} + b + \frac{1}{x^{2}} = 0 \) Rearrange terms

\( x^{2} + \frac{1}{x^{2}} + ax + \frac{a}{x} + b = 0 \) Add \( 2 \) and Subtract \( 2 \)

\( \left ( x^{2} + 2 + \frac{1}{x^{2}} \right ) + \left ( ax + \frac{a}{x} \right ) + b - 2 = 0 \) Factor each set of parentheses

\( \left ( x + \frac{1}{x} \right )^{2} + a \left ( x + \frac{1}{x} \right ) + b - 2 = 0 \)

Let \( v = x + \frac{1}{x} \), then we have \( v^{2} + av + b - 2 = 0 \).

Use the quadratic formula in attempt to solve for \( v \):
\( v = \frac{ (-a) \pm \sqrt{(-a)^{2}-4(1)(b - 2)} }{2(1)} \) Simplify

\( v = \frac{ -a \pm \sqrt{a^{2}-4b + 8} }{2} \)

Recall that \( v=x + \frac{1}{x} \), then multiplying by \( x \) yields:
\( vx = x^{2} + 1 \) Subtract \( vx \) from both sides

\( x^{2} - vx + 1 = 0 \)

In order for this to have "at least one real root", then the discriminant \( \Delta \geq 0 \):
\( (-v)^{2} - 4(1)(1) \geq 0 \) Simplify

\( v^{2} - 4 \geq 0 \) Add \( 4 \) to both sides

\( v^{2} \geq 4 \)

Since \( v = \frac{ -a \pm \sqrt{a^{2}-4b + 8} }{2} \) and \( v^{2} \geq 4 \), then:
\( \left ( \frac{ -a \pm \sqrt{a^{2}-4b + 8} }{2} \right )^{2} \geq 4 \) Simplify

\( \frac{ \left ( -a \pm \sqrt{a^{2}-4b + 8} \right )^{2} }{4} \geq 4 \) Multiply by \( 4 \)

\( \left ( -a \pm \sqrt{a^{2}-4b + 8} \right )^{2} \geq 16 \) Expand the left side

\( a^{2} \pm 2a \sqrt{a^{2}-4b + 8} + a^{2}-4b + 8 \geq 16 \) Combine like terms

\( 2a^{2} \pm 2a \sqrt{a^{2}-4b + 8} - 4b + 8 \geq 16 \) Divide both sides by \( 2 \)

\( a^{2} \pm a \sqrt{a^{2}-4b + 8} - 2b + 4 \geq 8 \) Isolate the square root term

\( \pm a \sqrt{a^{2}-4b + 8} \geq 8 - a^{2} + 2b - 4 \) Combine like terms

\( \pm a \sqrt{a^{2}-4b + 8} \geq 4 - a^{2} + 2b \) Square both sides

\( a^{2} \left ( a^{2}-4b + 8 \right ) \geq \left ( 4 - a^{2} + 2b \right )^{2} \) Expand both sides

\( a^{4} - 4a^{2}b + 8a^{2} \geq 16 - 4a^{2} + 8b - 4a^{2} + a^{4} - 2a^{2}b + 8b - 2a^{2}b
+4b^{2} \) Combine like terms

\( a^{4} - 4a^{2}b + 8a^{2} \geq 16 - 8a^{2} + 16b + a^{4} - 4a^{2}b + 4b^{2} \) Simplify/Cancel out terms

\( 8a^{2} \geq 16 - 8a^{2} + 16b + 4b^{2} \) add \( 8a^{2} \) to both sides

\( 16a^{2} \geq 16 + 16b + 4b^{2} \) Divide by \( 4 \) on both sides

\( 4a^{2} \geq 4 + 4b + b^{2} \) Add \( 4b^{2} \) to both sides and rewrite

\( 4a^{2} + 4b^{2} \geq 5b^{2} + 4b + 4 \) Factor out a \( 5 \) on the right and complete the square

\( 4a^{2} + 4b^{2} \geq 5 \left ( b^{2} + \frac{4}{5}b + \frac{4}{25} \right ) - \frac{4}{5} + 4 \) Factor and simplify

\( 4a^{2} + 4b^{2} \geq 5 \left ( b + \frac{2}{5} \right )^{2} + \frac{16}{5} \) Divide both sides by \( 4 \)

\( a^{2} + b^{2} \geq \frac{5}{4} \left ( b + \frac{2}{5} \right )^{2} + \frac{4}{5} \)

This implies that \( a^{2} + b^{2} \) has a minimum at the vertex of the parabola on the right side when \( b = - \frac{2}{5} \), so:

\( \therefore a^{2} + b^{2} = \frac{4}{5} \)