# [SOLVED]Find minimum a²+b²

#### anemone

##### MHB POTW Director
Staff member
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.

#### SquareOne

##### New member
Here's my attempt at a 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:

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