Solving Quadratic Equations and Applications

TL;DR

You'll learn how to find the values that make a quadratic equation true using different methods. These solutions represent key points, like where a parabola crosses the x-axis. Understanding this lets you solve real-world problems involving curves and optimal points.

1. The Mental Model

Think of a quadratic equation as describing a U-shaped curve (a parabola). Solving it means finding where this curve hits the ground (the x-axis). These points are important because they often represent break-even points, maximum heights, or optimal values in real situations.

2. The Core Material

A quadratic equation is any equation that can be rearranged into the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are numbers, and $a
eq 0$. The solutions (also called roots or zeros) are the values of $x$ that make the equation true. There can be zero, one, or two real solutions.

Methods for Solving Quadratic Equations

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1. Factoring

This is often the quickest method if it's applicable. You rewrite the quadratic expression as a product of two binomials.

Steps:
1. Set the equation to $0$.
2. Factor the quadratic expression.
3. Set each factor equal to $0$ and solve for $x$.

Example: Solve $x^2 + 5x + 6 = 0$
$(x+2)(x+3) = 0$
$x+2=0 \implies x=-2$
$x+3=0 \implies x=-3$
Solutions are $x = -2$ and $x = -3$.

2. Using the Quadratic Formula

The quadratic formula always works, even when factoring is difficult or impossible. It's $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Steps:
1. Ensure the equation is in the form $ax^2 + bx + c = 0$.
2. Identify $a$, $b$, and $c$.
3. Substitute these values into the formula and simplify.

Example: Solve $2x^2 - 3x - 5 = 0$
Here, $a=2$, $b=-3$, $c=-5$.
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$
$x = \frac{3 \pm \sqrt{9 - (-40)}}{4}$
$x = \frac{3 \pm \sqrt{49}}{4}$
$x = \frac{3 \pm 7}{4}$
So, $x = \frac{3+7}{4} = \frac{10}{4} = \frac{5}{2}$ and $x = \frac{3-7}{4} = \frac{-4}{4} = -1$.
Solutions are $x = \frac{5}{2}$ and $x = -1$.

3. Completing the Square

This method converts the quadratic into a "perfect square trinomial," which makes it easy to solve by taking the square root. It's also the method used to derive the quadratic formula.

Steps:
1. Move the constant term ($c$) to the right side of the equation.
2. Divide the entire equation by $a$ (if $a
eq 1$).
3. Take half of the $x$-term coefficient ($b/2$), square it $((b/2)^2)$, and add it to both sides of the equation.
4. Factor the left side as a perfect square and simplify the right side.
5. Take the square root of both sides and solve for $x$.

Example: Solve $x^2 + 6x - 7 = 0$
$x^2 + 6x = 7$
Half of $6$ is $3$, and $3^2$ is $9$. Add $9$ to both sides:
$x^2 + 6x + 9 = 7 + 9$
$(x+3)^2 = 16$
$\sqrt{(x+3)^2} = \pm\sqrt{16}$
$x+3 = \pm 4$
$x+3=4 \implies x=1$
$x+3=-4 \implies x=-7$
Solutions are $x = 1$ and $x = -7$.

When to Use Which Method

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graph TD
    Start["Quadratic Equation: ax^2 + bx + c = 0"] --> PreferFactoring{"Can it be easily factored?"}
    PreferFactoring -- Yes --> Factor["Factor into (x+p)(x+q)=0"]
    PreferFactoring -- No --> UseFormulaOrComplete{"Is 'b' even and 'a' is 1?"}
    Factor --> SolveFactors["Set each factor to 0 & solve"]
    UseFormulaOrComplete -- Yes --> CompleteSquare["Complete the Square"]
    UseFormulaOrComplete -- No --> QuadraticFormula["Use Quadratic Formula"]
    SolveFactors --> Solutions["Solutions for x"]
    CompleteSquare --> Solutions
    QuadraticFormula --> Solutions

Applications of Quadratic Equations

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Quadratic equations often model situations involving:
* Projectile motion: The height of a thrown object over time (e.g., $h(t) = -16t^2 + v_0t + h_0$).
* Area problems: Optimizing the area of a shape given constraints.
* Maximizing/Minimizing: Finding the maximum profit or minimum cost.
* Geometry: Relationships involving squares or rectangles.

When applying them, context matters. Negative solutions for time or distance are usually discarded.

3. Worked Example

A ball is thrown upwards from a 3-meter platform with an initial velocity of 14 m/s. The height $h$ (in meters) of the ball after $t$ seconds is given by the equation $h(t) = -5t^2 + 14t + 3$. When does the ball hit the ground?

To find when the ball hits the ground, we need to find the time $t$ when the height $h(t)$ is $0$.
So, we set the equation to $0$:
$-5t^2 + 14t + 3 = 0$

This doesn't look easily factorable, and the leading coefficient is negative, so let's use the quadratic formula.
First, multiply by $-1$ to make the leading coefficient positive (optional, but often makes calculations cleaner):
$5t^2 - 14t - 3 = 0$

Now, identify $a=5$, $b=-14$, $c=-3$.
Substitute into the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$t = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(5)(-3)}}{2(5)}$
$t = \frac{14 \pm \sqrt{196 - (-60)}}{10}$
$t = \frac{14 \pm \sqrt{196 + 60}}{10}$
$t = \frac{14 \pm \sqrt{256}}{10}$
$t = \frac{14 \pm 16}{10}$

We have two possible solutions for $t$:
1. $t = \frac{14 + 16}{10} = \frac{30}{10} = 3$
2. $t = \frac{14 - 16}{10} = \frac{-2}{10} = -0.2$

Since time cannot be negative in this context, we discard $t = -0.2$.
The ball hits the ground after $3$ seconds.

4. Key Takeaways

• Quadratic equations describe parabolas and have up to two real solutions.
• Factoring is fast when applicable, but the quadratic formula always works.
• Completing the square is useful for understanding the structure of quadratics and deriving the formula.
• Always set the quadratic equation to $0$ before solving by factoring or using the quadratic formula.
• The discriminant ($b^2 - 4ac$) tells you the nature of the solutions: positive means two real, zero means one real, negative means no real solutions.

Common mistakes to avoid:
- Forgetting the $\pm$ in the quadratic formula or when taking square roots.
- Incorrectly identifying $a$, $b$, and $c$ (especially signs).
- Making arithmetic errors, particularly with negative numbers under the square root.
- Not checking if your solutions make sense in a real-world application (e.g., negative time).

5. Now Try It

Solve the equation $3x^2 - 10x + 8 = 0$ using any method you prefer. After finding the solutions, quickly check one of them by plugging it back into the original equation to see if it makes the equation true.

Success looks like: You correctly identify $x=2$ and $x=\frac{4}{3}$ as the solutions.