What is an Equation – Math Open Reference
In mathematics, an equation is a statement that two expressions are equal. It consists of two expressions on the left and right sides of the “equals” symbol. There are different math equations. Below is an example of a math equation,
7+8=15
The equation states that 7 + 8 is equal to 15. The most common math equations include one or more variables (symbols that represent unknowns). For example,
x=15+25
Since both sides of the equations must be equal, x=37. There is only one value of x that makes the equation true. This means 37 satisfies the equation. The way of finding the value(s) of a variable that satisfies an equation is called “solving the equation”.
It is a common mistake to think that an expression is an equation like
x^2+y^2-7.
Expressions do not assert equality. Remember, an equation must have an equal sign and it must be true.
Common Types of Algebraic Math Equations
There are a lot of types of mathematical equations. Algebraic, transcendental, parametric, differential, integral, are a few of the different types of equations. This article will focus on solving the most common algebraic equations.
1. Linear Math Equations
Linear equations are any math equations that can be written in the form
ax+b=0
where a and b are real numbers and x is a variable. This is sometimes called the standard form of a linear equation.
There are a lot of ways in solving linear equations, depending on the level of difficulty of the linear equation you are trying to solve. The rule of thumb in solving any equation is balancing. Whatever you do to one side of an equation, should also do it to the other side. This will lead us to the properties of equality that will be heavily used in solving math equations. For example, solve
9(x+4)+2x=64-3x
First, simplify both sides of the equation. Clear out any groupings by distributive law. Then combine like terms.
\begin{alignedat}{2}9x+36+2x=64−3x\\ 11x+36=64-3x\end{alignedat}
The next step is to add 3x to both sides and subtract 36 from both sides of the equation.
11x+3x+36-36=64-36-3x+3x\\14x=28
Then divide both sides by 14.
\frac { 14x }{ 14 } =\frac { 28 }{ 24 }\\x=2
In a separate article, we will deal with other linear equations such as those with fractional coefficients and proportion equations.
2. Quadratic Math Equations
Quadratic equations are any equation that can be arranged in standard form
ax^{ 2 }+bx+c=0 , where a≠0
There are several ways to solve quadratic equations like factoring, square root property, the quadratic formula by completing the square, and the recently devised method last 2019 by Professor Loh. In this article, we will discuss the three basic ways—factoring, square root property, and the quadratic formula. The quadratic formula and square root property will always work. While the factoring method won’t always work. The latter needs a little more work to do for it to be used in some quadratics.
Solving by Factoring
Some quadratics can be solved by just factoring them. In this method, we will need to remember the Zero Product Property which states
If ab=0, then a=0 or b=0 (or both).
This means that if \left( x-2 \right) \left( x+3 \right) =0 , then
x-2=0 or x+3=0
Therefore, x=2 or x=3 . Zero product property won’t work like
ab=10 , then a=2 or b=5.
The equation must always be equal to zero for it to work.
To solve using factoring, we want all the terms to be on one side, and zero on the other side. As mentioned above, the zero product property won’t work if no side is equal to zero. In other words, the quadratic equation must always be in standard form. For more details on how to factor polynomials, read the article for Factoring Polynomials. For example, solve
{ x }^{ 2 }-x=0
Since the equation is in standard form, we can factor out x on the left. Then use the zero product property to solve for the values of x .
x\left( x-1 \right) =0
x=0 or x=1
Now, let’s try solving { x }^{ 2 }-4x=45 . First, write the equation in standard form by subtracting 45 from both sides.
\begin{alignedat}{2}{ x }^{ 2 }+4&x-45=45-45\\{ x }^{ 2 }+4&x-45=0\end{alignedat}
Then, factor the left side and use the zero product property to solve for x .
\left( x+9 \right) \left( x-5 \right) =0
x+9=0 or x-5=0
x=-9 or x=5
Factoring works provided the quadratic can be easily factored, which is not always the case.
Square Root Property
The Square Root Property tells that
if { x }^{ 2 }\quad =\quad a , then x=\pm \sqrt { a }
The \pm symbol is read as “plus or minus” which means exactly how it’s read. One is x=\sqrt { a } and the other is x=-\sqrt { a } . This property is always used in the Quadratic Formula and when completing the square. It can only be used when taking the square root of both sides of the equation that can be done to solve for the variable.
Example 1: Solve { x }^{ 2 }-289=0
First, add 289 to both sides, then take the square root of both sides.
\begin{alignedat}{2}{ x }^{ 2 }-289+289&=0+289\\\sqrt { { x }^{ 2 } } &=\sqrt { 289 }\\x&=\pm 17\end{alignedat}
Therefore, x can be either 17 or -17 .
Example 2: Solve { \left( 3x-6 \right) }^{ 2 }-225=0
Add 225 to both sides then take the square root.
\begin{alignedat}{2}(3x-6)^{ 2 }-225+225&=0+225\\\sqrt { { \left( 3x-6 \right) }^{ 2 } } &=\sqrt { 225 }\\3x-6&=\pm 15\end{alignedat}
We have 3x-6=15 and 3x-6=-15. Solve both equations to get x=7 or x=-3.
Example 3: Solve { \left( x+4 \right) }^{ 2 }+8=0.
Subtract 8 from both sides then take the square root.
\begin{alignedat}{2} { \left( x+4 \right) }^{ 2 }+8-8&=0-8\\ \sqrt { { \left( x+4 \right) }^{ 2 } } &=\sqrt { -8 }\\ x&=-4\pm 2i\sqrt { 2 } \end{alignedat}
Subtract 4 from both sides to get x=-4\pm 2i\sqrt { 2 }. We have complex solutions to the equation.
Quadratic Formula
The quadratic formula is derived by completing the square of the standard form of any quadratic equation. We will show you how to derive the formula in different articles but for now, here’s the formula.
x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }
This formula always works for any quadratic equation you want to solve. You just have to make the equation in standard form.
Example: Solve { x }^{ 2 }-5x-7=0.
Since the equation is already in standard form, let’s proceed to identify the values of a, b and c.
a=1, b=-5 and c=-7.
Plugin the values of the variables into the formula and solve for x.
\begin{alignedat}{2} x&=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }\\x&=\frac { -\left( -5 \right) \pm \sqrt { { \left( -5 \right) }^{ 2 }-4\left( 1 \right) \left( -7 \right) } }{ 2\left( 1 \right) }\\x&=\frac { 5\pm \sqrt { 25+28 } }{ 2 }\\x&=\frac { 5\pm \sqrt { 53 } }{ 2 } \end{alignedat}
You must know at least two ways to solve quadratic equations so may use one to double-check your answer.
3. Polynomial Equations
Polynomial equations take away the limit of the exponent you can have in an equation. This means that linear, quadratic, and cubic equations are all polynomial equations. Higher-degree polynomial equations can sometimes be solved by factoring.
Example 1: Solve { x }^{ 3 }+16{ x }^{ 2 }+64x-361x=0.
The polynomial equation looks daunting, but if we carefully pay attention, we can manipulate the left side to reveal a pattern.
First, factor out x from the left side.
x\left( { x }^{ 2 }+16x+64-361 \right) =0
It’s tempting to subtract 361 from 64 but the expression is actually a “difference of squares”. The { x }^{ 2 }+16x+64 and 361 are both perfect squares. Factor out the entire expression using what we know about the difference of squares.
x\left( { x+8-19 } \right) \left( x+8+19 \right) =0
Simplify the left side then solve forx.
x\left( { x-11 } \right) \left( x+27 \right) =0
x=0, x=11 or x=-27
It is very helpful to master factoring different levels of polynomials to help you solve polynomial equations. But polynomial equations won’t always be easy to solve using factorization. Some may need more than factorization like the Rational Root Theorem.
4. Radical Equations
Radical equations are any equation with a variable inside a radical. In this article, we will focus on solving square root equations. Remember, always isolate the square root expression on one side.
Example 1: Solve \sqrt { x+3 } -9=0
First, add 9 to both sides to isolate \sqrt { x+3 }. Then square both sides to cancel the square root and solve for x.
\begin{alignedat}{2}\sqrt { x+3 } -9+9&=0+9\\{ \left( \sqrt { x+3 } \right) }^{ 2 }\quad &=\quad { 9 }^{ 2 }\\x+3&=81\\x+3-3&=81-3\\x&=78\end{alignedat}.
As to any equation, we always want to plug back the value of the variable we obtained into the equation to check whether or not our solution is right. We do this especially to radical math equations because we can sometimes have an extraneous solution—an erroneous value of the variable that doesn’t satisfy the equation.
\begin{alignedat}{2}\sqrt { 78+3 } -9& ≟0\\\sqrt { 81 } -9& ≟0\\9-9&≟0\\0&=0\quad True\end{alignedat}
Most of the time, we will be dealing with quadratic equations when solving square root equations.
Example 2: \sqrt { 5-x } -1=x
First, isolate \sqrt { 5-x } by adding 1 to both sides, then square both sides.
\begin{alignedat}{2}\sqrt { 5-x } -1+1&=x+1\\{ \left( \sqrt { 5-x } \right) }^{ 2 }&={ \left( x+1 \right) }^{ 2 }\\5-x&=x^2+2x+1\end{alignedat}
Arrange the equation to standard form.
\begin{alignedat}{2}x^2+2x+x+1-5&=0\\x^2+3x-4&=0\end{alignedat}
We can solve the quadratic equation by factoring.
\begin{alignedat}{2}x^2+3x-4&=0\\\left( x+4 \right) \left( x-1 \right) &=0 \end{alignedat}
\begin{alignedat}{2}x&=-4\end{alignedat} or \begin{alignedat}{2}x&=1\end{alignedat}
Check for extraneous solution(s).
\begin{alignedat}{2}\sqrt { 5-\left( -4 \right) } -1+1&≟-4+1\\\sqrt { 9 } -1+1&≟-3\\3-1+1 &≟ -3\\3&\neq -3\quad False\end{alignedat}
\begin{alignedat}{2}\sqrt { 5-1 } -1+1 &≟ 1+1\\\sqrt { 4-1 } +1&≟2\\2-1+1&≟ 2\\2&=2\quad True\end{alignedat}
Therefore, the only solution to the equation is x=1.
Keep in mind that you always have to check for extraneous solution(s) when solving square root equations. A square root equation may have only one solution or even none at all.
5. Exponential Equations
Exponential equations are equations in which variables occur as an exponent. There are two ways in solving exponential equations. But we will restrict ourselves to solving exponential equations by making the bases on both sides of the equations the same (the first thing we want to check first if possible). We will discuss solving exponential equations using logs in another article.
Example 1: { 2 }^{ x }=16
First, let’s turn the right side of the equation into an exponential expression with the same base as the left side. 16 can be expressed as { 2 }^{ 4 }.
{ 2 }^{ x }={ 2 }^{ 4 }
Once both sides have the same base, we can equate their exponents to solve for x.
{ 2 }^{ x }={ 2 }^{ 4 }\\x=4
Example 2: 16^{ x-5 }\cdot { \left( \frac { 1 }{ 8 } \right) }^{ -x-\frac { 1 }{ 6 } }=\sqrt { 2 }
The equation looks a little complicated but it’s not. Just make all the bases the same using exponent properties.
\begin{alignedat}{2}16^{ x-5 }\cdot { \left( \frac { 1 }{ 8 } \right) }^{ -x-\frac { 1 }{ 6 } }&=\sqrt { 2 }\\2^{ 4\left( x-5 \right) }\cdot { 2 }^{ -3\left( -x-\frac { 1 }{ 6 } \right) }&=2\frac { 1 }{ 2 }\\ { 2 }^{ 4x-20 }\cdot { 2 }^{ 3x+2 }&={ 2 }^{ \frac { 1 }{ 2 } }\end{alignedat}
Simplify the left side of the equation.
\begin{alignedat}{2}2^{ 4x-20+3x+\frac { 1 }{ 2 } }&={ 2 }^{ \frac { 1 }{ 2 } }\\2^{ 7x-\frac { 39 }{ 2 } }&={ 2 }^{ \frac { 1 }{ 2 } }\end{alignedat}
Equate the equations from both sides and solve for x.
\begin{alignedat}{2}7x-\frac { 39 }{ 2 } &=\frac { 1 }{ 2 }\\2\left( 7x-\frac { 39 }{ 2 } \right)& =\frac { 1 }{ 2 } \left( 2 \right)\\14x-39&=1\\14x-39+39&=1+39\\\frac { 14x }{ 14 } &=\frac { 40 }{ 14 }\\x&=\frac { 20 }{ 7 }\end{alignedat}
Conclusion
Remember that when solving an equation, you should always apply the rule “whatever you do to one side, do it to the other.” In the next articles, we will go over more algebraic equations and give easy ways to solve them. In the meantime, browse our blog page for different fun educational math blogs.
