Lesson 1: Extraction of Roots

New Concept

We're leveling up from linear to quadratic equations! This lesson introduces the 'extraction of roots' method, a powerful technique to solve equations like $ax^2 + c = 0$ by isolating the squared variable and taking its square root.

What’s next

Next, you'll master this method through interactive examples and practice cards. Then, tackle challenge problems involving geometry and even compound interest!

Property

A quadratic equation can be written in the standard form

where $a$, $b$, and $c$ are constants, and $a$ is not zero.

Examples

• The equation $2x^2 + 5x - 3 = 0$ is a quadratic equation in standard form, where $a=2$, $b=5$, and $c=-3$.
• The equation $4x^2 = 8x - 1$ is also quadratic. It can be written in standard form as $4x^2 - 8x + 1 = 0$.
• The equation $5x - 10 = 0$ is not quadratic because it is a first-degree equation; the highest power of $x$ is 1.

Explanation

Unlike linear equations which create straight lines, quadratic (or second-degree) equations have a variable squared. This $x^2$ term creates a U-shaped curve called a parabola. The key is that the highest exponent on the variable is 2.

Property

To solve a quadratic equation of the form

1. Isolate $x^2$ on one side of the equation.
2. Take the square root of each side.

Remember that every positive number has two square roots, one positive and one negative.

Examples

• To solve $3x^2 = 75$, first divide by 3 to get $x^2 = 25$. Then take the square root of both sides to find $x = ±\sqrt{25}$, so the solutions are $x=5$ and $x=-5$.
• For $2x^2 - 18 = 0$, add 18 to both sides to get $2x^2 = 18$. Divide by 2 to get $x^2=9$. The solutions are $x = ±3$.
• To solve $x^2 - 15 = 0$, first isolate $x^2$ to get $x^2 = 15$. The exact solutions are $x = ±\sqrt{15}$.

Explanation

Extraction of roots is a method to 'undo' a squared variable. Think of it as working backward. First, get the $x^2$ term by itself. Then, take the square root of both sides to find the value of $x$, making sure to include both positive and negative roots ($\±$).

Property

To solve a formula for a variable that is squared, use the extraction of roots method. Isolate the squared variable on one side of the equation, and then take the square root of both sides. For the volume of a cone, $V = \frac{1}{3}\pi r^2 h$, we can solve for the radius $r$.

Because a physical dimension like radius must be positive, we typically use only the positive square root.

Examples

• To solve the Pythagorean theorem $a^2 + b^2 = c^2$ for side $a$, isolate $a^2$ to get $a^2 = c^2 - b^2$. The formula for $a$ is $a = \sqrt{c^2 - b^2}$.
• The formula for the surface area of a sphere is $S = 4\pi r^2$. To solve for the radius $r$, divide by $4\pi$ to get $r^2 = \frac{S}{4\pi}$. The formula for $r$ is $r = \sqrt{\frac{S}{4\pi}}$.
• The formula for kinetic energy is $E = \frac{1}{2}mv^2$. To solve for velocity $v$, multiply by 2 and divide by $m$ to get $v^2 = \frac{2E}{m}$. The formula for $v$ is $v = \sqrt{\frac{2E}{m}}$.

Explanation

This technique lets you rearrange a scientific or geometric formula to find a specific quantity. By solving for a variable like radius or time before plugging in numbers, you create a more direct and reusable formula for your specific problem.

Property

We can use extraction of roots to solve quadratic equations of the form

We start by isolating the squared expression, $(x - p)^2$.

Examples

• To solve $5(x - 3)^2 = 125$, first divide by 5 to get $(x - 3)^2 = 25$. Take the square root: $x - 3 = ±5$. This gives two equations: $x-3=5$ (so $x=8$) and $x-3=-5$ (so $x=-2$).
• To solve $2(x+4)^2 - 8 = 0$, add 8 and divide by 2 to get $(x+4)^2=4$. Take the square root: $x+4 = ±2$. The solutions are $x=-2$ and $x=-6$.
• In $-3(x+1)^2 = -75$, divide by -3 to get $(x+1)^2=25$. Take the square root: $x+1 = ±5$. The solutions are $x=4$ and $x=-6$.

Explanation

This method extends extraction of roots to cases where an entire expression in parentheses is squared. The strategy is the same: treat the squared parenthesis as a single block, isolate it, take the square root of both sides, and then solve for $x$.

Property

After $n$ years, the amount of money, $A$, in an account is given by the formula

where $P$ is the original principal and $r$ is the interest rate compounded annually. When solving for the rate $r$ over two years ($n=2$), we solve the equation $A = P(1+r)^2$ using extraction of roots.

Examples

• To find the rate needed for 2000 dollars to grow to 2205 dollars in 2 years, solve $2205 = 2000(1+r)^2$. This gives $(1+r)^2 = 1.1025$, so $1+r = 1.05$. The rate is $r=0.05$ or 5%.
• The price of a collectible inflated from 400 dollars to 484 dollars in two years. The inflation rate $r$ is found by solving $484 = 400(1+r)^2$. This gives $(1+r)^2 = 1.21$, so $1+r=1.1$. The rate is $r=0.10$ or 10%.
• If a 5000 dollars investment grows to 5408 dollars in two years, we solve $5408 = 5000(1+r)^2$. This gives $(1+r)^2 = 1.0816$, so $1+r=1.04$. The rate is $r=0.04$ or 4%.

Explanation

This formula models how money grows with compound interest. To find the interest rate needed to reach a financial goal in two years, you can treat $(1+r)$ as the unknown block in the equation and solve for it using extraction of roots.