Quadratic equations by factoring

Algebra, grade 9 Lesson 12 of 26Learning

Quadratic equations by factoring

An equation in which the unknown appears squared – like $\displaystyle {x^{2}-5x+6}={0}$ – is a quadratic equation (an equation of the second degree). Unlike an equation of the first degree, such as $\displaystyle {2x+3}={7}$ , it can have two solutions. One way to solve it: factor, and use one simple fact about zero.

A product equal to zero

This works only with 0. From

\displaystyle {(x-2)(x+7)}={8}

you cannot conclude that

\displaystyle {x-2}={8}

If a product of two numbers is 0, at least one of them is 0: $\displaystyle {a\cdot b}={0}$ means $\displaystyle {a}={0}$ or $\displaystyle {b}={0}$ . So in the equation $\displaystyle {(x-2)(x+7)}={0}$ either $\displaystyle {x-2}={0}$ , and then $\displaystyle {x}={2}$ , or $\displaystyle {x+7}={0}$ , and then $\displaystyle {x}={-7}$ .

The equation has two solutions: $\displaystyle {x}={2}$ and $\displaystyle {x}={-7}$ . Check: $\displaystyle {(2-2)(2+7)}={0\cdot9}={0}$ , and $\displaystyle {(-7-2)(-7+7)}={-9\cdot0}={0}$ .

Exercise 1

Solve the equations: set each factor equal to zero.

How many solutions does this equation have?

An equation with a common factor

Do not divide both sides by

\displaystyle x

: that loses the solution

\displaystyle {x}={0}

When one side is 0 and the other is not yet a product, factor it first. In $\displaystyle {x^{2}-5x}={0}$ the common factor is $\displaystyle x$ :

\begin{aligned}x^{2}-5x&=0\\ x(x-5)&=0\end{aligned}

$\displaystyle {x}={0}$ or $\displaystyle {x-5}={0}$ , so $\displaystyle {x}={0}$ or $\displaystyle {x}={5}$ .

An equation with a difference of squares

In $\displaystyle {x^{2}}={9}$ move the 9 to the side of $\displaystyle x^{2}$ , and factor by the formula $\displaystyle {a^{2}-b^{2}}={(a-b)(a+b)}$ :

\begin{aligned}x^{2}-9&=0\\ (x-3)(x+3)&=0\end{aligned}

$\displaystyle {x}={3}$ or $\displaystyle {x}={-3}$ – two numbers whose square is 9. And $\displaystyle {x^{2}}={-4}$ has no solution at all: the square of a number is never negative.

An equation with a trinomial

In $\displaystyle {x^{2}-5x+6}={0}$ look for two numbers whose sum is $\displaystyle -5$ and whose product is $\displaystyle 6$ : $\displaystyle -2$ and $\displaystyle -3$ .

\begin{aligned}x^{2}-5x+6&=0\\ (x-2)(x-3)&=0\end{aligned}

$\displaystyle {x}={2}$ or $\displaystyle {x}={3}$ . Check $\displaystyle {x}={2}$ : $\displaystyle {2^{2}-5\cdot2+6}={4-10+6}={0}$ – true.

Moving every term to one side

The fact about zero works only when one side is 0. So first move all the terms to one side, and only then factor:

\begin{aligned}x^{2}&=3x+10 &&/-3x-10\\ x^{2}-3x-10&=0\\ (x-5)(x+2)&=0\end{aligned}

$\displaystyle {x}={5}$ or $\displaystyle {x}={-2}$ .

Solving by factoring: the steps

Move all the terms to one side, so the other side is 0.
Factor (a common factor, a trinomial, a formula).
Set each factor equal to zero and solve.
Check by substituting in the original equation.

A quadratic equation has at most two solutions; sometimes one, sometimes none.

Exercise 2

Factor the first side, then solve.

Exercise 3

First move all the terms to one side.
Then factor and solve.

Exercise 4

A number is multiplied by the number that is 3 greater than it, and the product is 40.
Write an equation, solve it, and find all the numbers that fit.

Formulas: algebra, grade 9

Open as a page

The laws of powers

Multiplying powers with the same base
$\displaystyle a^{m}\cdot a^{n}$ $\displaystyle =$ $\displaystyle a^{m+n}$
Dividing powers with the same base
$\displaystyle a^{m}:a^{n}$ $\displaystyle =$ $\displaystyle a^{m-n}$ $\displaystyle a\neq0$
A power of a power
$\displaystyle \left(a^{m}\right)^{n}$ $\displaystyle =$ $\displaystyle a^{m\cdot n}$
A power of a product
$\displaystyle (a\cdot b)^{n}$ $\displaystyle =$ $\displaystyle a^{n}\cdot b^{n}$
A power of a quotient
$\displaystyle \left(\frac{a}{b}\right)^{n}$ $\displaystyle =$ $\displaystyle \frac{a^{n}}{b^{n}}$ $\displaystyle b\neq0$
The power zero
$\displaystyle a^{0}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle a\neq0$

The laws of square roots

The root of a product
$\displaystyle \sqrt{a\cdot b}$ $\displaystyle =$ $\displaystyle \sqrt{a}\cdot\sqrt{b}$ $\displaystyle a\ge0,\ b\ge0$
The root of a quotient
$\displaystyle \sqrt{\frac{a}{b}}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{a}}{\sqrt{b}}$ $\displaystyle a\ge0,\ b>0$
The square of a root
$\displaystyle \left(\sqrt{a}\right)^{2}$ $\displaystyle =$ $\displaystyle a$ $\displaystyle a\ge0$
The root of a square
$\displaystyle \sqrt{a^{2}}$ $\displaystyle =$ $\displaystyle a$ $\displaystyle a\ge0$

The extended distributive law and the short multiplication formulas

The extended distributive law
$\displaystyle (a+b)(c+d)$ $\displaystyle =$ $\displaystyle ac+ad+bc+bd$
The square of a sum
$\displaystyle (a+b)^{2}$ $\displaystyle =$ $\displaystyle a^{2}+2ab+b^{2}$
The square of a difference
$\displaystyle (a-b)^{2}$ $\displaystyle =$ $\displaystyle a^{2}-2ab+b^{2}$
The difference of squares
$\displaystyle (a+b)(a-b)$ $\displaystyle =$ $\displaystyle a^{2}-b^{2}$

The parabola

Standard form
$\displaystyle y=ax^{2}+bx+c$
Vertex form – the vertex is $\displaystyle (b,k)$
$\displaystyle y=a(x-b)^{2}+k$
Product form – it meets the $\displaystyle x$ -axis at $\displaystyle x_1$ and $\displaystyle x_2$
$\displaystyle y=a(x-x_1)(x-x_2)$
The vertex’s $\displaystyle x$ , in standard form
$\displaystyle x=-\frac{b}{2a}$

The quadratic equation

The roots of $\displaystyle {ax^{2}+bx+c}={0}$ , when $\displaystyle a\neq0$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$