# Solving quadratics by factorising

## How to solve quadratic equations by factorising

A quadratic equation is a polynomial equation of degree of order $$2$$. $$ax^{2}+bx+c$$ is the general form of a quadratic polynomial. If we equate this polynomial to zero, we get a quadratic equation.** **The general form of a quadratic equation is $$ax^{2}+ bx+c=0$$.

Here, $$x$$ represents an unknown variable, while $$a$$, $$b$$, and $$c$$ represent the known numbers. Here, $$a$$, $$b$$, and $$c$$ are the numerical coefficients of the quadratic equation.

**Solution of a quadratic equation**

The solution of a quadratic equation is the value of an unknown variable $$x$$, which satisfies the given quadratic equation. The solution of a quadratic equation is called zeroes or roots of the equation. In a quadratic equation, the equation has two roots or zeroes because it has a degree of order $$2$$.

**E2.5: Derive and solve quadratic equation by factorisation.**

**
The factorisation method**

Quadratic equations can be solved by many methods: The factorisation method is one of them. In the factorisation method, first, factorise the quadratic polynomial in the given equation, then put each factor equal to zero, and find the roots of the equation accordingly.

Let us take an example and learn how this method works. Suppose an organisation wants to build a big hall of carpet area of $$300\;\text{square meters}$$, with its length one meter more than twice its breadth. What should be the length and breadth of the hall?

Let the breadth of the hall be $$x\;\text{meters}$$.

Then its length should be $$\left (2x+1 \right)\;\text{meters}$$.

Now, the area of the hall is $$\left (2x+1 \right) \times x =\left (2x^ {2} +x \right) = 2x^{2} +x$$.

So, the breadth of the hall should satisfy the equation:

$$2x^ {2} +x=300$$

$$2x^{2}+x-300$$

This is a quadratic equation.

Now we need to factorise the above equation to find the value of $$x$$.

$$2x^{2}+x-300=0$$

$$2x^{2}+25x-24x-300=0$$

$$( 2x+25)-12\left ( 2x+25\right )$$

$$(2x+25)(x-12)=0$$.

From the above equation, we get $$2x+25=0$$ and $$x-12=0$$.

Then, $$x-12=0$$ gives $$x=12$$ and $$2x+25=0$$ gives $$x=-\frac{25}{2}$$. But $$x=-\frac{25}{2}$$ cannot be considered, because length does not take negative values.

Therefore, the breadth of the hall is $$12\;\text{meters}$$, and the length of the hall is $$\left ( 2\times 12+1 \right ) \;\text{meter}=25\;\text{meters}$$.

**Worked examples of using the factorisation method**

**Example 1: **Find the root of the equation $$x^ {2} +5x+6=0$$ by factorisation.

**Step 1: Factorise the given quadratic equation by splitting the middle term.**

$$x^{2} +5x+6=x^ {2} +\left (3x+2x \right) +6$$

$$x^{2} +3x+2x+6 = 0$$

$$x\left (x+3 \right) +2\left (x+3 \right)$$

$$=\left (x+3 \right) \left (x+2 \right)$$

**Step 2: Put each factor equal to zero.**

$$x+3=0$$ or $$x+2=0$$

Now, $$x+3=0$$ gives $$x=-3$$, and $$x+2=0$$ gives $$x=-2$$.

**Step 3: Answer in the preferred notation.**

So, $$x=-3$$ and $$x=-2$$ are the roots of the given equation.

**Example 2: **By using factorisation, solve the quadratic equation $$2x^{2}-6x+4= 0$$.

**Step 1: Divide both the sides of the equation by 2.**

$$x^ {2}-3x+2= 0$$

**Step 2: Factorise the given quadratic equation by splitting the middle term.**

$$x^{2}-3x+2=x^ {2}-2x-x+2$$

$$x\left (x-2 \right)-1\left (x-2 \right) = 0$$

$$\left (x-2 \right) \left (x-1 \right)=0$$

$$\left (x-2 \right) \left (x-1 \right)=0$$

**Step 3: Put each factor equal to zero.**

$$x-2=0$$ or $$x-1=0$$

Now, $$x-2=0$$ gives $$x=2$$, and $$x-1=0$$ gives $$x=1$$.

**Step 4: Answer in the preferred notation.**

Therefore, $$x=2$$ and $$x=1$$ are the roots of the given quadratic equation.

**Example 3**: Find the root of the quadratic equation $$\left (x+2 \right)^{2}-9=0$$ by factorisation.

**Step 1: Simplify the equation to change it into the general form.**

$$\left (x+2 \right)^{2}-9 =x^ {2}+4x+4-9$$

$$x^ {2} +4x-5 = 0$$

**Step 2: Factorise the given quadratic equation by splitting the middle term.**

$$x^ {2} +4x-5=x^ {2} +5x-x-5$$

$$x\left (x+5 \right)-\left (x+5 \right)=0$$

$$\left (x+5 \right) \left (x-1 \right)=0$$

**Step 3: Put each factor equal to zero.**

$$x+5=0$$ or $$ x-1=0$$

Now, $$x+5=0$$ gives $$x=-5$$, and $$x-1=0$$ gives $$x=1$$.

**Step 4: Answer in preferred notation.**

$$x=-5$$ or $$x=$$1 are the roots of the given quadratic equation.