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

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In summary, the purpose of finding the minimum value of $a^2+b^2$ in a quadratic equation is to determine the lowest possible value of the function, which can be used to solve real-world problems. To find the minimum value, we can use the vertex form of a quadratic equation. The minimum value can be negative, depending on the context of the problem. It is not always unique, but if there is no vertical line of symmetry, it will be. The value of a can affect the minimum value, with a positive a resulting in the lowest possible value and a negative a resulting in the highest possible value.
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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.
 
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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:

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

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

1. What is the minimum value of $a^2+b^2$ in a quadratic equation?

The minimum value of $a^2+b^2$ in a quadratic equation is 0. This occurs when $a$ and $b$ are both equal to 0.

2. How do you find the minimum value of $a^2+b^2$ in a quadratic equation?

The minimum value of $a^2+b^2$ in a quadratic equation can be found by completing the square or by using the vertex form of a quadratic equation.

3. Can the minimum value of $a^2+b^2$ be negative?

No, the minimum value of $a^2+b^2$ cannot be negative. Since both $a$ and $b$ are squared, their values will always be positive or 0, resulting in a minimum value of 0.

4. How does the value of $a^2+b^2$ affect the shape of a quadratic graph?

The value of $a^2+b^2$ affects the shape of a quadratic graph by determining the location of the vertex. A larger value of $a^2+b^2$ will result in a wider and flatter parabola, while a smaller value will result in a narrower and taller parabola.

5. Are there any real-life applications of finding the minimum value of $a^2+b^2$ in a quadratic equation?

Yes, finding the minimum value of $a^2+b^2$ in a quadratic equation is useful in optimization problems, such as finding the minimum cost or maximum profit in a business scenario. It can also be applied in physics to determine the minimum energy or force required for a certain task.

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