[Common Math 1 Part 13] Quadratic Equations and the Quadratic Formula

In the previous post, we introduced complex numbers and organized their operations. Now we return to quadratic equations and write their solutions in a general form.

Derive the quadratic formula by completing the square, so we can solve quadratic equations even when factoring does not work immediately.

Let us fix the flow first.

• Some quadratic equations are easy to solve by factoring.
• But not every quadratic equation factors neatly.
• So we need one method that works in general.
• The result of organizing that method is the quadratic formula.

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1. Why Do We Need the Quadratic Formula?

For example,

$$x^2-5x+6=0$$

factors as

$$(x-2)(x-3)=0,$$

so we immediately get $x=2$ and $x=3$.

But now consider

$$2x^2-3x+7=0.$$

At that point, factoring is no longer something we can read off comfortably.

So the real question is this:

Can we always solve a quadratic equation, even when factoring is not obvious?

The quadratic formula is the answer to that question.

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2. Start from the General Form

The general quadratic equation is

$$ax^2+bx+c=0 \qquad (a\neq 0).$$

We require $a\neq 0$ because if $a=0$, then the $x^2$ term disappears and the equation is no longer quadratic. In this post, we work within the real and complex number systems.

So our goal is clear:

Express the solutions of $ax^2+bx+c=0$ in terms of $a$, $b$, and $c$.

The key method is completing the square.

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3. Deriving the Quadratic Formula by Completing the Square

Start with

$$ax^2+bx+c=0.$$

Since $a\neq 0$, divide both sides by $a$:

$$x^2+\frac{b}{a}x+\frac{c}{a}=0.$$

Move the constant term to the right:

$$x^2+\frac{b}{a}x=-\frac{c}{a}.$$

Now we want the left side to become a perfect square.

3-1. What Should We Add?

In general, to turn

$$x^2+px$$

into a perfect square, we add

$$\left(\frac{p}{2}\right)^2,$$

because

$$\left(x+\frac{p}{2}\right)^2 =x^2+2\cdot x\cdot \frac{p}{2}+\left(\frac{p}{2}\right)^2 =x^2+px+\left(\frac{p}{2}\right)^2.$$

Here $p=\frac{b}{a}$, so we add

$$\left(\frac{b}{2a}\right)^2$$

to both sides:

$$x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2.$$

The left side becomes

$$\left(x+\frac{b}{2a}\right)^2.$$

On the right side, we can first rewrite

$$-\frac{c}{a}=-\frac{4ac}{4a^2}.$$

Then using a common denominator gives

$$\frac{-4ac+b^2}{4a^2}=\frac{b^2-4ac}{4a^2}.$$

So we obtain

$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$$

3-2. Take Square Roots and Isolate $x$

Taking square roots on both sides gives two possibilities, so the symbol $\pm$ appears:

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

Now move $\frac{b}{2a}$ to the right:

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

That is the quadratic formula.

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4. What Does the Formula Tell Us?

The formula

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

shows the solutions of a quadratic equation all at once.

The part we should especially notice is

$$b^2-4ac.$$

This quantity is called the discriminant.

Depending on whether this value is positive, zero, or negative, the shape of the roots changes. In a short preview:

• if the discriminant is positive, there are two distinct real roots
• if it is 0, there is one repeated real root
• if it is negative, there are two complex conjugate roots

We will study that part carefully in the next post.

For now, two points matter most:

• the roots of a quadratic equation can be expressed generally in terms of $a$, $b$, and $c$
• even when a real-number solution stops working, the formula can continue into complex numbers

For example, in $x^2-4x+5=0$, the discriminant is negative, so the roots are not real but complex: $2\pm i$.

This is easier to feel once you actually change the coefficients and watch the discriminant, roots, and graph behavior move together. As you try it, pay special attention to how the roots change when the discriminant becomes positive, zero, or negative.

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5. Examples Make the Meaning Clearer

5-1. A Case That Also Factors

Consider

$$x^2-5x+6=0.$$

Here,

$$a=1, \qquad b=-5, \qquad c=6.$$

Substitute into the formula:

$$x=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot6}}{2\cdot1}.$$

So

$$x=\frac{5\pm\sqrt{25-24}}{2}=\frac{5\pm1}{2},$$

which gives

$$x=3, \qquad x=2.$$

Even when factoring already works, the quadratic formula gives the same answer. So the formula is not a trick for rare cases. It is the general method that covers the whole topic.

5-2. A Case with Complex Roots

Now consider

$$x^2-4x+5=0.$$

Then

$$a=1, \qquad b=-4, \qquad c=5,$$

so

$$x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot5}}{2\cdot1}.$$

Simplifying gives

$$x=\frac{4\pm\sqrt{16-20}}{2}=\frac{4\pm\sqrt{-4}}{2}.$$

In the complex number system,

$$\sqrt{-4}=\sqrt{4\cdot(-1)}=2\sqrt{-1}=2i,$$

so

$$x=\frac{4\pm2i}{2}=2\pm i.$$

This shows that the quadratic formula is not only for real roots. It organizes the full set of roots, including complex ones.

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6. Common Mistakes

The formula is powerful, but substitution errors happen often.

6-1. Losing the Sign of $b$

The formula uses $-b$. So if $b=-5$, then $-b=5$. If we lose that sign, the whole answer changes.

6-2. Squaring $b$ Incorrectly

If $b=-4$, then

$$b^2=(-4)^2=16.$$

We must square the whole number, not just attach a minus sign carelessly.

6-3. Forgetting That $2a$ Divides the Entire Numerator

In

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$

the denominator $2a$ divides both terms in the numerator.

For example,

$$\frac{4\pm2i}{2}$$

becomes $2\pm i$, not $4\pm i$.

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7. Key Takeaways

• The general quadratic equation is $ax^2+bx+c=0$ with $a\neq 0$.
• When factoring is not obvious, we build a general method by completing the square.
• That method produces $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
• The quadratic formula covers both factorable cases and cases with complex roots.
• The expression $b^2-4ac$ is especially important because it controls the nature of the roots.

In the next post, we will study exactly what $b^2-4ac$ tells us, that is, how the discriminant organizes the number and type of roots.

One-line conclusion:

The quadratic formula is not a backup trick for when factoring fails. It is the general method that organizes the roots of every quadratic equation at once.