Lesson 110: Using the Quadratic Formula

New Concept

For the quadratic equation $ax^2 + bx + c = 0$,

What’s next

Next, you will apply this powerful formula to solve different types of quadratic equations, including those found in real-world scenarios.

Property

For the quadratic equation $ax^2 + bx + c = 0$,

Explanation

It's the ultimate 'cheat code' for solving any quadratic equation. Derived from completing the square, this formula saves you the trouble. Just identify your $a$, $b$, and $c$ values from the standard form of the equation, plug them in, and chug out the answers. It’s a reliable shortcut that always works!

Examples

To solve $x^2 - 7x + 12 = 0$, use $a=1, b=-7, c=12$: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} = \frac{7 \pm 1}{2}$, so $x=4$ and $x=3$.
To solve $2x^2 + 5x - 4 = 0$, use $a=2, b=5, c=-4$: $x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-4)}}{2(2)} = \frac{-5 \pm \sqrt{57}}{4}$.
To solve $3x^2 + 2x + 5 = 0$, use $a=3, b=2, c=5$: $x = \frac{-2 \pm \sqrt{2^2 - 4(3)(5)}}{6} = \frac{-2 \pm \sqrt{-56}}{6}$, which has no real solution.

Property

A quadratic equation is an equation whose graph is a parabola. The standard form is $ax^2 + bx + c = 0$.

Explanation

Think of it as any equation where the variable's highest power is two, creating a U-shaped parabola graph. Before using the quadratic formula, you must arrange the equation into its standard form, $ax^2 + bx + c = 0$. This ensures you correctly identify the $a$, $b$, and $c$ coefficients for the formula.

Examples

The equation $-18x + x^2 = -32$ is rearranged into standard form as $x^2 - 18x + 32 = 0$.
The equation $16h^2 + 25 = 40h$ is rearranged into standard form as $16h^2 - 40h + 25 = 0$.
The equation $x^2 + 80 = 21x$ is rearranged into standard form as $x^2 - 21x + 80 = 0$.

Sometimes an equation needs a little tidying up before the quadratic formula can work its magic. This example will cover the key idea of rearranging equations.

Example Problem

Use the quadratic formula to solve $x^2 - 11x = -24$ for $x$.

Step-by-Step

1. First, we must rearrange the equation into the standard form $ax^2 + bx + c = 0$.

2. Now we identify our coefficients: $a=1$, $b=-11$, and $c=24$.
3. Apply the quadratic formula.

4. Substitute the values for $a$, $b$, and $c$.

5. Simplify the expression inside and outside the square root.

6. Find the two possible solutions for $x$.

The solutions are $x = 8$ and $x = 3$.

Property

The height of an object tossed upwards can be modeled by $-4.9t^2 + vt + s = 0$, where $t$ is time, $v$ is initial velocity, and $s$ is initial height.

Explanation

This formula is a real-world quadratic equation that models the path of a tossed object. It helps predict when something will hit the ground. Since time moves forward, we only use the positive solution from the quadratic formula. A negative answer for time doesn't make sense, so we discard it as an impossible solution.

Examples

A ball is tossed from a height of 60 meters with a velocity of 8 m/s. Use $-4.9t^2 + 8t + 60 = 0$ to find when it lands. The positive solution is $t \approx 4.41$ seconds.
A ball is tossed from a 50-meter cliff with a velocity of 6 m/s. Use $-4.9t^2 + 6t + 50 = 0$ to find when it lands. The positive solution is $t \approx 3.82$ seconds.