Let $f(x)$ be an expression in $x$. For example, $f(x)=3x+2$ or $f(x)=8x^{2}-\sqrt{x}+5$. Where the graph of the equation $y=f(x)$ crosses the $y$-axis is called the $y$-intercepts and where it crosses the $x$-axis is called the $x$-intercept.
To find the $x$-intercept of the graph of $y=f(x)$, we set $y=0$ and solve for $x$. Therefore, the $x$-intercepts of the graph of $y=f(x)$ are solutions to $f(x)=0$ (also called the roots of $f$).
Similarly to find the $y$-intercept, we substitute $x=0$ in the equation $y=f(x)$ and solve for $y$.
Linear equations
An equations that can be written of the form
\[ax+b=0,\qquad(a\neq0)\]
where $a$ and $b$ are (fixed) real numbers and $x$ is the variable (or the unknown) is called a linear equation. To solve the equation, move $b$ to the other side
\[ax=-b\]
and then move the coefficient $a$ to the denominator of $b$; that is,
\[x=-\frac{b}{a}.\]
(In fact we subtract $b$ from both sides and then divide both sides of the resulting equation by $a$).
- The solution of $ax+b=0$ is the $x$-intercept of the straight line
$y=ax+b$ (the $x$-intercept is where the graph intersects the $x$-axis).
Power equations
Consider an equation of the form
\[x^{n}=a\]
where $a$ is a fixed number and $n$ is an integer. Then the solution(s) of the above the equation are:
\[x=\begin{cases}
\sqrt[n]{a} & \text{$n$ is odd)}\\
\pm\sqrt[n]{a} & \text{ $n$ is even and $a>0$}\\
\text{no real solution} & \text {$n$ is even and $a<0$}
\end{cases}\]
Quadratic equations
The general form of quadratic equations (or a second-degree equation) is
\[ax^{2}+bx+c=0,\quad(a\neq0)\tag{i}\]
For example,
\[3x^{2}-5x+2=0\]
is a quadratic equation. In this example, $a=3,b=-5$, and $c=2$.
- The solutions (also called roots or zeros) of $ax^{2}+bx+c=0$ are the $x$-intercepts of the graph $y=ax^{2}+bx+c$. The graph of $y=ax^{2}+bx+c$ opens upward if $a>0$ and downward if $a<0$.
Quadratic formula
The solutions (or roots) of the quadratic equation $ax^2+bx+c=0$ can always be found using the following general formula, called the quadratic formula:
\[ \bbox[#F2F2F2,5px,border:2px solid black]{\large x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}}\]
When we apply the general formula to the equation $3x^{2}-5x+2=0$ with $a=3,b=-5$, and $c=2$, we get:
\begin{align*}
x & =\frac{5\pm\sqrt{5^{2}-4\cdot3\cdot2}}{2\cdot3}\\
& =\frac{5\pm1}{6}\\
& =1\quad\text{or}\quad\frac{2}{3}
\end{align*}
The part of the quadratic formula that is under the square root, $D=b^{2}-4ac$, is called the discriminant of the equation $ax^{2}+bx+c=0$. There are three cases:
\[D=b^2-4ac\]
See the following figures. \[ \bbox[#F2F2F2,5px,border:2px solid black]{\large ax^{2}+bx+c=a(x-r_{1})(x-r_{2}).}\]
For example, because the roots of $3x^{2}-5x+2$ are 1 and $2/3$, we have \[ \bbox[#F2F2F2,5px,border:2px solid black]{r_{1}+r_{2}=-\frac{b}{a},\qquad r_{1}r_{2}=\frac{c}{a}}\]
Because if we expand $a(x-r_{1})(x-r_{2})$, we have For example, the roots of $3x^{2}-5x+2=0$ are $1$ and $2/3$ and If we can rewrite the quadratic equation $ax^{2}+bx+c=0$ as A binomial $x^{2}+bx$ becomes a perfect square if we add $\left(\frac{b}{2}\right)^{2}$ to it: Suppose $a=1$: To solve \begin{align*} Take the square roots of both sides: \begin{align*} Move $-b/2$ to the right hand side: \begin{align*} For example, let us solve The left hand side is a complete square: \[(x-3)^{2} =14\]
Take the square root of each side \begin{align*} The solutions therefore are $3-\sqrt{14}$ and $3+\sqrt{14}$. In summary: \[h=\frac{b}{2a},\quad\text{and}\quad k=c-ah^2=c-\frac{b^2}{4c}\]
One of the applications of completing the square is in finding the center of a circle when its equation is expressed in the general form, as illustrated in the following equation. By completing the square, we can find the maximum or minimum value of a quadratic polynomial.
(a) $D>0, a>0$, two real roots
(b) $D>0, a<0$, two real roots
(c) $D=0, a>0$, double root
(d) $D<0, a>0$, no real roots
\begin{align*}
3x^{2}-5x+2 & =3\left(x-1\right)\left(x-\frac{2}{3}\right)\\
& =(x-1)(3x-2)
\end{align*}
\[a(x-r_{1})(x-r_{2})=ax^{2}-a(r_{1}+r_{2})x+ar_{1}r_{2}.\]
So by comparing with $ax^{2}+bx+c$, we realize that $-a(r+r_{2})=b$ and $ar_{1}r_{2}=c$.
\[1+\frac{2}{3}=-\frac{b}{a}=-\left(-\frac{5}{3}\right),\qquad1\cdot\frac{2}{3}=\frac{c}{a}=\frac{2}{3}\]
\[a\times1^{2}+b\times1+c=a+b+c=0\]
\[a\cdot(-1)^{2}+b\cdot(-1)+c=a-b+c=b-b=0.\]
Completing the square
\[(x+A)^{2}=C,\]
then we can easily solve it by taking the square roots of both sides (see here). The left-hand side is a perfect square— the square of $x+A$, and if we expand it, we get
\[ (x+A)^{2}=x^{2}+2Ax+A^{2}. \]
So in a perfect square, the constant term is the square of half of the coefficient of $x$.
\[x^{2}+bx+\left(\frac{b}{2}\right)^{2}=\left(x+\frac{b}{2}\right)^{2} \]
This method, which is called completing the square, has many applications in different parts of mathematics. One of these applications is in solving the quadratic equations and deriving the quadratic formula we have seen in this section.
\[x^{2}+bx+c=0\]
we add $(b/2)^{2}$ to both sides of the equation, so
\[ x^{2}+bx+\left(\frac{b}{2}\right)^{2}+c=\left(\frac{b}{2}\right)^{2} \]
Therefore, the sum of the first three terms makes a perfect square
\left(x+\frac{b}{2}\right)^{2}+c & =\left(\frac{b}{2}\right)^{2}\\
\left(x+\frac{b}{2}\right)^{2} & =\left(\frac{b}{2}\right)^{2}-c
\end{align*}
x+\frac{b}{2} & =\pm\sqrt{\left(\frac{b}{2}\right)^2-c}
\end{align*}
x & =-\frac{b}{2}\pm\sqrt{\left(\frac{b}{2}\right)^{2}-c}
\end{align*}Show an example
Hide the example
\[x^{2}-6x-5=0.\]
First, transpose 5; that is, add 5 to both sides
\[x^{2}-6x=5.\]
If now we add $(-6/2)^{2}=9$ to both sides, the left hand side will be a perfect square
\[x^{2}-6x+9=14\]
x-3 & =\pm\sqrt{14}\\
x & =3\pm\sqrt{14}
\end{align*}
Transformation of the equations of circles to the standard form by completing the square
Finding the maximum or minimum value of a quadratic polynomial
hence has a maximum.
\begin{align*}
y & =-\frac{1}{2}x^{2}+x+1\\
& =-\frac{1}{2}\left[x^{2}-2x-2\right]\\
& =-\frac{1}{2}\left[x^{2}-2x+(-1)^{2}-1-2\right]\\
& =-\frac{1}{2}\left[\left(x-1\right)^{2}-1-2\right]\\
& =-\frac{1}{2}\left[\left(x-1\right)^{2}-3\right]\\
& =-\frac{1}{2}(x-1)^{2}+\frac{3}{2}
\end{align*}
Because $-\frac{1}{2}(x-1)^{2}$ can never be positive, and it is zero when $x=1$, $y$ has a maximum of $3/2$ when $x=1$. The graph of $y=-\frac{1}{2}x^{2}+x+1$ is sketched in the following figure.
