Solution of simple quadratic equations

Solving equations and inequalities: Quadratic equations

Solution of simple quadratic equations

A quadratic equation has 0, 1, or 2 solutions. We give an example for each of the three cases .

The equation $x^2+1=0$ has no real solutions. Because a square is always greater than or equal to $0$ , the left-hand side is for each choice of $x$ greater than or equal to $1$ .

The equation $x^2-2x+1=0$ has exactly one solution. Because the left-hand side of the equation can be written as $(x-1)^2$ and this square can only be equal to $0$ if $x-1=0$ , there is only one solution, namely $x=1$ .

The equation $x^2-4=0$ has two solutions, namely $x=-2$ of $x=2$ .

The last example can also be understood as follows: the left-hand side of the equation, that is, $x^2-4$ , can be written as $(x+2)(x-2)$ and this product can only be equal to $0$ if one of the terms in the product is equal to $0$ . So $x+2=0$ or $x-2=0$ , that is, $x=-2$ or $x=2$ .

We have applied the following general rule.

$A\cdot B = 0\quad\text{if and only if }\quad A=0\;\;\text{or}\;\;B=0$ Here "or" must be understood such that $A$ and $B$ can also be both equal to $0$ .

In formal mathematical language, this rule is:

$A\cdot B = 0\quad\iff\quad A=0\;\;\lor\;\;B=0$

When a quadratic equation is in factored form you can easily read of the solutions. We give another example.

Let $a$ and $b$ be real numbers with $a\neq0$ .

$\begin{aligned} ax^2+bx=0&\implies x\cdot (ax+b)=0\\ \\ &\implies x=0\;\;\text{or}\;\;ax+b=0\\ \\&\implies x=0\;\;\text{or}\;\;x=-\frac{b}{a}\end{aligned}$

In this section we will discuss three methods to determine the solutions of a quadratic equation:

Completing the square
Factorisation by inspection
The quadratic formula

But before we are going to do this we look first at another simple quadratic equation, namely of the form $x^2=c$ with a given number $c$ . The number of solutions of this equation depends on the sign of $c$ .

Example: x² = c

Two solutions

If $c>0$ , then the equation

$x^2=c$ has two solutions, namely

$x=\sqrt{c}\quad\vee\quad x=-\sqrt{c}\text.$ The parabola and the horizontal line $y=c$ intersect in two points.

GeoGebra

One solution

If $c=0$ , then the equation

$x^2=c$ has one solution, namely

$x=0\text.$

The parabola and the horizontal line $y=0$ have one common point, namely $(0,0)$ .

GeoGebra

No solution

If $c=0$ , then the equation

$x^2=c$ has no solution

The parabola and the horizontal line do not intersect $y=c$

GeoGebra

Determine the exact solution of

$2x^2=32$ in $x$ by reduction.

$\begin{aligned} 2x^2=32 &\implies x^2=\frac{32}{2}=16=4^2\\ &\qquad\blue{\text{isolation of }x^2\text{ and the right-hand side also written as a square}}\\ &\implies x=4\quad\lor\quad x=-4\\ &\qquad\blue{\text{taking square roots on both sides}}\end{aligned}$

New example

Mathcentre video

Solving Quadratic Equations (50:19)