Algebra is a branch of mathematics that deals with the expressions and equations. An expression is said to be a combination of constants and variables connected with the help of four algebraic operations: + (add), - (subtract), ^ (exponentiation) and x (multiplication). Few example of expressions are :

4a + 3b,

8x$^{3}$,

p$^{2}$ + 2q$^{3}$,

x$^{11}$ - 4xy + y$^{9}$ + 14.

When an expression involves equal sign (=), it is referred as an equation. It means that an equation has two expressions, each on either side of equal sign. Left expression is called left hand side or LHS, while right one is eventually termed as right hand side or RHS. The above expression can be written as equations as shown below :

4a = - 3b,

8x$^{3}$ = 0,

p$^{2}$ = - 2q$^{3}$,

x$^{11}$ - 4xy + y$^{9}$ = - 14.

The expressions as well as equations may contain any number of terms. In this article, we are going to learn about the trinomial equations and methods of solving them.

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In algebra, an expression with 3 terms is called a trinomial. The word "trinomial" can be split into two words "tri" which means "three" and "nomial" meaning "terms". Hence, the name trinomial is justified. If the trinomial expression is equated to zero, it is called a trinomial equation. We can say that a polynomial equation which involves three terms is known as a trinomial equation. In general, the suchl equation acquires the following form -

General Trinomial Equation:** x = p + x$^{n}$**

where, x is a variable. This form is discovered in 18th century by mathematician Johann Heinrich Lambert.

**For example -**p + 3q + 8r with 3 variables p, q and r.

$3x^{3} - 3y + 9z^2$ with 3 variables x, y and z.

3mn + 9m + 5n with 2 variables m and n.

In algebra, we often refer a these equations as quadratic equations. It can be of the form ax$^2$ + bx + c = 0. Here the highest degree of the equation should be 2. To have a trinomial equation as quadratic, a $\neq$ 0.

The trinomial equations have different methods of solving according to the format. They can be solved by factoring common terms out in case of having any common factor.**For example -**

6x$^{2}$ + 3xy

= 3x (2 + y)

Generally, a trinomial quadratic equation does have specific methods of solving them. These equations are commonly found in mathematics. **A quadratic equation can be solved by the following methods according to the need.**

From the quadratic equation ax$^2$ + bx + c=0

__We need to find two things:__

(**i**) Product = a $\times$ c

(**ii**) Sum = b

After knowing these two values, find two numbers that satisfies these two conditions. this will help you to break the middle term.

Then we can factorize by combining two two terms together. That can be evaluated separately to get the values of x.

It is good to know which is the right method to apply for a given trinomial equations.

Now let us try to apply those methods to solve a given trinomial as follows in the examples.

Below you could see solving trinomial equations

**Example 1:** Solve the following trinomial equation: 3x$^2$ – 8x – 3 = 0.

**Solution:** Given: 3x$^2$ – 8x – 3 = 0 → (1)

Here let us apply factorization method. For that we need to do the following:

Let us compare (1) with ax$^2$ + bx + c = 0.

a = 3, b = -8, c = - 3.

Need to find two things:

(1). Product = a $\times$ c = 3 $\times$ – 3 = - 9.

(2). Sum = b = - 8.

Now find two numbers so that their product is – 9 and their sum is – 8.

They are – 9 and 1; and that is – 9x and 1x.

Therefore (1)$\Rightarrow$ 3x$^2$ – 8x – 3 = 0.

$\Rightarrow$ 3x$^2$ – 9x + x – 3 = 0.

$\Rightarrow$ (3x$^2$ – 9x) + (x – 3) = 0 $\Rightarrow$ 3x (x – 3) +1 (x – 3) = 0.

$\Rightarrow$ (x – 3) (3x + 1) = 0.

$\Rightarrow$ x = 3, x = $\frac{1}{-3}$.

Therefore The solution set is {3, $\frac{-1}{3}$}.

**Example 2:** Solve the following trinomial equation x$^2$ – x – 6 = 0.

**Solution:** Given: x$^2$ – x – 6 = 0 → (1)

Here let us apply the method of completion of squaring method.

(1) $\Rightarrow$ (x – $\frac{1}{2}$)$^2$ – ($\frac{1}{2}$ )$^2$ – 6 = 0 [Here divide the coefficient of x by 2 and subtract

$\Rightarrow$ (x – $\frac{1}{2}$ )$^2$ – [$\frac{1}{4}$ + 6] = 0 the square of that value to form a square].

$\Rightarrow$ (x – $\frac{1}{2}$ )$^2$ – $[\frac{1 + 24}{4}]$ = 0.

$\Rightarrow$ (x – $\frac{1}{2}$ )$^2$ = $\frac{25}{4}$

$\Rightarrow$ x – $\frac{1}{2}$ = $\pm$ $\sqrt{\frac{25}{4}}$ = $\pm$ $\frac{5}{2}$ .

Therefore x = $\frac{1}{2}$$\pm$$\frac{5}{2}$$\Rightarrow$ x = $\frac{1}{2}$ + $\frac{5}{2}$, $\frac{1}{2}$ - $\frac{5}{2}$.

$\Rightarrow$ x = $\frac{[1\ +\ 5]}{2}$, x = $\frac{[1\ -\ 5]}{2}$.

$\Rightarrow$ x = 3, - 2.

**Example 3:** Solve the following trinomial equation x$^2$ – 6x + 8 = 0.

**Solution:**

Here let us apply the formula method to solve the given equation.

Here a = 1, b = -6, c = 8.

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

= $\frac{-(-6) \pm \sqrt{(-6)^2\ -\ 4(1)(8)}}{2(1)}$

= $\frac{6 \pm \sqrt{36\ -\ 32}}{2}$

= $\frac{6 \pm 2}{2}$

Therefore, x = $\frac{6\ +\ 2}{2}$ , $\frac{6\ -\ 2}{2}$

$\Rightarrow$ x = 4, 2.

**(1)** Solve: x$^2$ – 3x – 4 = 0.

[**Ans**: x = {4, -1}]

**(2)** Solve: 2x$^2$ + 7x + 5 = 0

[**Ans**: x = {-1, - $\frac{5}{2}$ }].

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