2.3 Solving quadratic equations

Quadratic equations appear when a variable is squared. Areas, trajectories, and many optimization questions produce them. Solving one means finding every input that makes the equation true. Unlike a usual linear equation, a quadratic can have two, one, or no real solutions.

This is foundational to intermediate algebra. Be comfortable with distribution and factoring from Module 2.1 and balancing equations from Module 2.2. OpenStax's quadratic-equations lesson gives further worked practice.

A basic idea: where does the expression reach zero?

An equation such as $x^2-5x+6=0$ is quadratic because the highest power of $x$ is $2$. If we can write the left side as $(x-2)(x-3)$, the product is zero when either factor is zero. Thus $x=2$ and $x=3$ are its roots. On the graph of $y=x^2-5x+6$, they are the points where the curve meets the horizontal axis.

The square-root symbol has one nonnegative value, but solving $x^2=9$ gives both $x=3$ and $x=-3$. That is why a $\pm$ appears when we take square roots while solving.

A quick tour of methods

Factor when the product is simple

First move all terms to one side so the other side is zero. For $x^2+5x+6=0$, find two numbers that multiply to $6$ and add to $5$: they are $2$ and $3$. Then $(x+2)(x+3)=0$, so $x=-2$ or $x=-3$. If the leading coefficient is not $1$, look for a common factor first and verify a proposed factorization by expanding it.

Complete the square when factoring is awkward

The identity $(x+d)^2=x^2+2dx+d^2$ tells us how to turn $x^2+bx$ into a square: take half of $b$, then square it. To solve $x^2+4x-1=0$, move the constant and add the same $4$ to both sides:

$$x^2+4x=1 \quad\Longrightarrow\quad x^2+4x+4=5 \quad\Longrightarrow\quad (x+2)^2=5.$$

Taking both square-root possibilities gives $x+2=\pm\sqrt5$, so $x=-2\pm\sqrt5$. The new $4$ is not guessed: it is $(4/2)^2$. The diagram shows why adding that small corner makes a full square.

The picture uses positive side lengths to explain the identity. The algebraic identity remains true for every real $x$.

Use the general formula and count real roots

For $ax^2+bx+c=0$ with $a\ne0$, completing the square gives

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

The number $D=b^2-4ac$ is the discriminant. If $D>0$, there are two distinct real roots; if $D=0$, one repeated real root; if $D<0$, no real roots. The formula always applies, but factoring or completing the square can explain a problem more clearly. Check roots in the original equation, especially after moving or dividing terms.

Application 1: the sides of a rectangle

A rectangle has area $30$ square units. Its length is one unit more than its width. Find both sides.

1. Name the width. Let it be $x>0$; then the length is $x+1$, and $x(x+1)=30$.
2. Move to zero and factor. $x^2+x-30=(x+6)(x-5)=0$, giving $x=-6$ or $x=5$.
3. Apply the meaning and check. A width cannot be negative, so $x=5$. The sides are $5$ and $6$, and their area is $30$.

Both numbers solve the algebraic equation, but only one describes a rectangle. Module 2.6 studies such restrictions.

Application 2: an equation that does not factor neatly

Solve $x^2+4x-1=0$ over the real numbers.

1. Move the constant. $x^2+4x=1$.
2. Complete the square. Half of $4$ is $2$, and $2^2=4$. Add $4$ to both sides: $(x+2)^2=5$.
3. Take both square roots. $x=-2+\sqrt5$ or $x=-2-\sqrt5$. Substitution confirms each: if $x+2=\pm\sqrt5$, then $(x+2)^2=5$, exactly the rearranged equation.

The solutions are approximately $0.236$ and $-4.236$. Rounding early would lose accuracy, so keep the exact square-root form until the last step.