Solving Quadratic Equations by Factoring and the Quadratic Formula

Hello. In the previous lesson, you solved linear inequalities and represented whole solution sets precisely. Quadratic equations return us to equality, but with a new feature: the highest power of the variable is $2$. A quadratic can have two, one, or no real-number solutions.

This lesson develops two complementary methods. Factoring is fast when the quadratic has a convenient product structure; the quadratic formula is the reliable general method. Work through every algebraic line by hand. The goal is not merely to memorize a formula, but to understand why each method is valid and how to choose between them.

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1. What a quadratic equation is

A quadratic equation can be written in standard form as

$$a{x}^2+bx+c=0,$$

where $a$, $b$, and $c$ are real numbers and

$$a\ne 0.$$

The condition $a\ne 0$ matters: if $a=0$, there is no $x^2$ term and the equation is linear instead.

For example,

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

is already in standard form, with

$$a=3,b=-7,c=2.$$

An equation need not begin in standard form. Consider

$$2x^2+5=7x.$$

Subtract $7x$ from both sides before deciding how to solve it:

$$2x^2-7x+5=0.$$

This “zero on one side” form is essential for factoring. It lets us turn a quadratic into a product equal to zero.

The zero-product property

The key fact behind factoring is the zero-product property:

If $u$ and $v$ are real numbers and $uv=0$, then $u=0$, $v=0$, or both.

Why? If $u\ne 0$, then division by $u$ is allowed:

$$uv=0$$ $$v=\frac{0}{u}=0.$$

So a product can be zero only when at least one factor is zero. This apparently simple fact turns a difficult-looking quadratic equation into two linear equations.

Do not apply the property unless the entire left side is a product and the other side is exactly zero. For instance,

$$(x-2)(x+5)=0$$

does permit the zero-product property. But

$$(x-2)(x+5)=7$$

does not.

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2. Solving by factoring

Factoring reverses expansion. If you can rewrite a quadratic as two linear factors, solve each factor separately.

A monic quadratic: leading coefficient $1$

Solve

$$x^2-2x-15=0.$$

We seek two numbers that have:

• product $-15$, the constant term;
• sum $-2$, the coefficient of $x$.

Those numbers are $-5$ and $3$. Therefore,

$$x^2-2x-15=(x-5)(x+3).$$

The equation becomes

$$(x-5)(x+3)=0.$$

By the zero-product property, either factor must be zero:

$$x-5=0$$

or

$$x+3=0.$$

Thus,

$$x=5$$

or

$$x=-3.$$

A quick check confirms both values:

$$5^2-2(5)-15=25-10-15=0,$$ $$(-3{)}^2-2(-3)-15=9+6-15=0.$$

Two useful patterns

Before using a general factoring method, look for simpler structure.

First, factor out a greatest common factor if one exists.

$$3x^2-75=0$$ $$3(x^2-25)=0.$$

Since $3\ne 0$, the important factor is $x^2-25$:

$$x^2-25=(x-5)(x+5).$$

So the solutions are

$$x=-5,x=5.$$

Second, recognize a difference of squares.

$$9x^2-64=0.$$

Both terms are squares:

$$9x^2=(3x{)}^2,64=8^2.$$

Hence

$$9x^2-64=(3x-8)(3x+8).$$

Set each factor equal to zero:

$$3x-8=0,3x+8=0.$$

The solutions are

$$x=\frac{8}{3},x=-\frac{8}{3}.$$

When the leading coefficient is not $1$

Now solve

$$6x^2-5x-4=0.$$

The simple “product $c$, sum $b$” rule is not enough because the leading coefficient is $6$, not $1$. Instead, multiply $a$ and $c$:

$$ac=6(-4)=-24.$$

Find two integers whose product is $-24$ and whose sum is $-5$. They are $3$ and $-8$. Split the middle term using these numbers:

$$6x^2-5x-4=0$$ $$6x^2+3x-8x-4=0.$$

Group the terms in pairs:

$$(6x^2+3x)+(-8x-4)=0.$$

Factor the greatest common factor from each group:

$$3x(2x+1)-4(2x+1)=0.$$

Now both terms contain the same binomial factor:

$$(2x+1)(3x-4)=0.$$

Apply the zero-product property:

$$2x+1=0$$

or

$$3x-4=0.$$

Therefore,

$$x=-\frac{1}{2}$$

or

$$x=\frac{4}{3}.$$

Why the $ac$ method works

This is worth understanding rather than treating as a trick. Suppose a quadratic factors as

$$(Ax+B)(Cx+D).$$

Expanding gives

$$(Ax+B)(Cx+D)=AC{x}^2+(AD+BC)x+BD.$$

So, compared with

$$a{x}^2+bx+c,$$

we have

$$a=AC,b=AD+BC,c=BD.$$

The two pieces of the middle coefficient are $AD$ and $BC$. Their sum is

$$AD+BC=b.$$

Their product is

$$(AD)(BC)=(AC)(BD)=ac.$$

That is exactly why we search for two numbers whose product is $ac$ and whose sum is $b$. Splitting the middle term recreates the pieces that came from expansion.

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3. The quadratic formula: a method that always applies

Factoring is efficient, but not every quadratic with integer coefficients factors neatly using integers. For example,

$$x^2+6x+4=0$$

does not factor into integer binomials. This does not mean it has no real solutions. It means we need a different method.

For every quadratic in standard form,

$$a{x}^2+bx+c=0,$$

the quadratic formula is

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

The symbol $\pm$ represents two calculations:

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

and

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

Applying the formula carefully

Solve

$$x^2+6x+4=0.$$

First identify the coefficients:

$$a=1,b=6,c=4.$$

Substitute them into the formula, keeping the substitution visually explicit:

$$x=\frac{-(6)\pm \sqrt{(6{)}^2-4(1)(4)}}{2(1)}.$$

Now simplify:

$$x=\frac{-6\pm \sqrt{36-16}}{2}$$ $$x=\frac{-6\pm \sqrt{20}}{2}.$$

Since

$$\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5},$$

we get

$$x=\frac{-6\pm 2\sqrt{5}}{2}.$$

Divide both parts of the numerator by $2$:

$$x=-3\pm \sqrt{5}.$$

Thus the two solutions are

$$x=-3+\sqrt{5}$$

and

$$x=-3-\sqrt{5}.$$

The values are real but irrational. Factoring over the integers could not reveal them, while the quadratic formula does.

Parentheses prevent sign errors

Suppose

$$x^2-4x+13=0.$$

Here,

$$a=1,b=-4,c=13.$$

Write the negative coefficient in parentheses:

$$x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1)(13)}}{2(1)}.$$

Then

$$x=\frac{4\pm \sqrt{16-52}}{2}$$ $$x=\frac{4\pm \sqrt{-36}}{2}.$$

Because a negative number has no real square root, this equation has no real solution. Later, when complex numbers are introduced, such equations will have complex solutions. For now, state the real-number conclusion clearly.

The discriminant

The expression under the square root,

$$b^2-4ac,$$

is called the discriminant. Denote it by

$$D=b^2-4ac.$$

It tells you the number of real solutions before you finish the formula.

Value of $D$	Real solutions
$D>0$	Two distinct real solutions
$D=0$	One real solution, repeated twice
$D<0$	No real solutions

For example, the discriminant of

$$x^2+6x+4=0$$

is

$$D=6^2-4(1)(4)=20>0,$$

so two distinct real solutions are expected.

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4. Where the quadratic formula comes from

The quadratic formula is not an isolated rule to memorize. It is the result of applying completing the square to the general quadratic equation.

Start with

$$a{x}^2+bx+c=0,$$

where $a\ne 0$. First move the constant term:

$$a{x}^2+bx=-c.$$

Divide all terms by $a$:

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

To turn the left side into a perfect square, add the square of half the coefficient of $x$. Half of $\frac{b}{a}$ is

$$\frac{b}{2a}.$$

Add its square to both sides:

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

The left side factors as a perfect square:

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

Use the common denominator $4a^2$ on the right:

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

Therefore,

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

Taking square roots gives

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

Because the $\pm$ already represents both possible signs, replacing $|a|$ with $a$ does not change the set of two possible solutions. Isolating $x$ and combining terms produces

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

The formula therefore packages the completing-the-square method into a single reusable result. The discriminant appears because it is the quantity that remains after completing the square; it determines whether a real square root is possible.

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5. Choosing a method and checking your work

A practical hand-calculation routine is:

1. Rewrite in standard form:

   $$a{x}^2+bx+c=0.$$

2. Factor out a greatest common factor, if there is one.

3. Look for a recognizable structure, especially a difference of squares.

4. If factor pairs appear manageable, factor and use the zero-product property.

5. If factoring is not evident, use the quadratic formula.

6. Check each proposed solution in the original equation when the arithmetic is manageable.

The methods are consistent with one another. For instance, the factored equation

$$(x-5)(x+3)=0$$

has roots $5$ and $-3$. Expanding it gives

$$x^2-2x-15=0.$$

Using the quadratic formula on that expanded equation produces the same two roots. Factoring is faster here because the structure is visible; the formula is valuable because it remains available when that structure is not visible.

A final distinction matters:

• “It does not factor nicely” means factoring is inconvenient or impossible over the integers.
• “It has no real solution” means the discriminant is negative.

These are completely different conclusions.

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Conclusion

You can now solve a quadratic equation in two major ways:

• Factoring: rewrite the equation with zero on one side, factor it into linear factors, then use the zero-product property.
• Quadratic formula: identify $a$, $b$, and $c$ carefully and compute

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

You also saw why the factoring-by-grouping method searches for numbers with product $ac$ and sum $b$, and why the quadratic formula arises from completing the square. The discriminant

$$b^2-4ac$$

tells you whether to expect two, one, or no real solutions.

Next, you will solve a pair of linear equations simultaneously using substitution and elimination.