# Solving Quadratic Equations Algebraically

## Introduction

Most of us first learn about linear equations in Secondary One, and quadratic equations in Secondary Two. What distinguishes a quadratic equation from a linear equation is the highest power of *x*, also known as the degree of an equation. The degree of a linear equation is 1 (the highest power of *x* is 1), while the degree of a quadratic equation is 2 (the highest power of *x* is 2).

While it is possible to solve a quadratic equation graphically, this article goes through the three methods, taught under Elementary Mathematics, to solve a quadratic equation **algebraically**.

## Solving a Linear Equation Algebraically

Before we go through the different methods to solve a quadratic equation, let us revisit the solving of a simple linear equation.

Given a general linear equation of the form* \(ax+b=c\) *where *a*, *b* and *c* are real numbers and* \(a \ne 0\),* solving the equation simply means finding the value of *x*. To do so, we **balance the equation** by performing a mathematical operation on both sides of the equation, one at a time.

**Step 1: Subtract ***b* from both sides of the equation

*b*from both sides of the equation

*\(ax+b=c\)*

* \(ax=c-b\)*

**Step 2: Divide by a on both sides of the equation**

**\(ax=c-b\)**

** \(x=\frac{c-b}{a}\) **

Therefore,* \(x=\frac{c-b}{a}\)* is the solution of the equation* \(ax+b=c.\)*

### Method 1: Solving a Quadratic Equation by Factorisation

Most of us first learn how to solve a quadratic equation by factorisation. This is often taught in Secondary Two.

To illustrate the steps involved in Method 1, we will use the quadratic equation* \(2x^2-5x-3=0.\)*

**Step 1:** Using the multiplication frame, factorise the quadratic expression* \(2x^2-5x-3\)*

**Step 2:** Express the quadratic expression *\(2x^2-5x-3\) *as a product of two linear factors

*\(2x^2+2-5x-3=0\)*

*\((2x+1)(x-3)=0\)*

**Step 3:** Using the Zero Product Principle, derive two separate linear equations

*\((2x+1)(x-3)=\) *0

*\(2x+1=0\)** *or* \(x-3=0\)*

**Step 4:** Solve each linear equation separately to obtain two values of *x*

*\(2x+1=0\)* or *\(x-3=0\)*

*\(x=-\frac{1}{2}\)* or* \(x=3\) *

Solving a quadratic equation by factorisation means that the final value(s) of *x* must be a rational number (either an integer or a fraction).

### Method 2: Solving a Quadratic Equation by Completing the Square

Subsequently, we are introduced to two new methods to solve a quadratic equation. This is often taught in Secondary Three. Unlike Method 1 where the final value(s) of *x* must be rational, Method 2 or 3 is often used when the final value(s) of *x* are irrational (cannot be expressed as a fraction).

The first of these two new methods introduced involves the concept of completing the square. Be reminded that completing the square may only be performed on a quadratic expression when its coefficient of *x*^{2} is 1.

To illustrate the steps involved in Method 2, we will use the quadratic equation *\(x^2+6x-10=0.\)*

**Step 1: ** Express *\(x^2+6x-10=0\)* in the completed square form *\((x+p)^2+q\)*

*\(x^2+6x-10=0\)*

*\((x+3)^2-3^2-10=0\)*

*\((x+3)^2-19=0\)*

**Step 2**: Balance the equation and make *x* the subject

*\((x+3)^2-19=0\)*

*\((x+3)^2=19\)*

*\(x+3=\pm\sqrt{19}\)*

*\(x=-3\pm\sqrt{19}\)*

**Step 3: ** Express the final answer(s) to the required degree of accuracy

*\(x=-3\pm\sqrt{19}\)*

*\(x=1.3588 \)* or \(-7.3588\)

*\(x \approx 1.36\)* or *\(-7.36\)* (3 s.f.)

What happens then if a quadratic equation has a coefficient of *x*^{2} that is not \(1\) (or \(0\))? This is where Method 3 comes in handy!

### Method 3: Solving a Quadratic Equation using the Quadratic Formula

If a quadratic equation has a coefficient of *x*^{2} that is not 1 (or 0), solving the quadratic equation using the Quadratic Formula is highly preferred over completing the square. Although it is still possible to solve the quadratic equation by completing the square, this scenario is beyond the scope of Elementary Mathematics.

Since the use of the Quadratic Formula is not bounded by certain conditions, it is more versatile and straightforward to use compared to completing the square. Hence, for both Elementary Mathematics and Additional Mathematics, we usually only complete the square when the question specifically requires a quadratic expression to be written in the completed square form.

According to the Quadratic Formula, the solution(s) to a quadratic equation of the form

*\(ax^2+bx+c=0,\) *where *a, b *and *c* are real numbers and* \(a \ne 0,\)* is *\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\)*

If you find the Quadratic Formula difficult to remember, you are in luck – the formula is included in the Formula Sheet provided during examinations.

To illustrate the steps involved in Method 3, we will use the quadratic equation* \(2x^2+3x-8=0.\)*

**Step 1: Determine the values of ***a*, *b* and *c*

*a*,

*b*and

*c*

For the quadratic expression *\(2x^2+3x-8,\)* *\(a=2\)*, *\(b=3\)* and *\(c=-8.\)*

**Step 2: Substitute the values of ***a*, *b* and *c* into the Quadratic Formula

*a*,

*b*and

*c*into the Quadratic Formula

Using the Quadratic Formula,

*\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)*

*\(x = {-(3) \pm \sqrt{(3)^2-4(2)(-8)} \over 2(2)}\)*

**Step 3: Express the final answer(s) to the required degree of accuracy**

*\(x = {-(3) \pm \sqrt{(3)^2-4(2)(-8)} \over 2(2)}\)*

*\(= \frac{-(3) \pm \sqrt{73}}{4}\)*

= \(1.386\) or \(-2.886\)

\(\approx 1.39\) or \(-2.89\) (3 s.f.)

## When to use which method?

Now that we have covered the three methods to solve quadratic equations algebraically, this flow chart below shows the thought process behind deciding which method to use.

Usually, when asked to solve a quadratic equation in an examination, the question will specifically mention if you are required to leave your answer(s) correct to a certain degree of accuracy.

Determining whether approximation is required for the final answer(s) posits the first important question in deciding which method to use – can the quadratic expression be factorised?

More often than not, if a degree of accuracy **is not specified**, the quadratic expression can be factorised and Method 1 (Multiplication Frame) should be considered.

On the other hand, if a degree of accuracy **is specified**, this means that the final answer should be approximated and factorisation is not possible. How do we then choose between Method 2 (Complete the Square) and Method 3 (Quadratic Formula)?

As mentioned earlier, completing the square may only be performed on a quadratic expression when its coefficient of * \(x^2\)* is \(1\). Furthermore, Method 2 (Complete the Square) should only be considered if the question specifically requires students to express the quadratic expression *\(x^2+bx+c\)* in the completed square form *\((x+p)^2+q.\)*

Otherwise, Method 3 (Quadratic Formula) is often more straightforward.

## Conclusion

In summary, do not just remember and stick to one method of solving quadratic equations algebraically. Depending on the equation given, a particular method of solving may be easier compared to the rest. Therefore, learning when to identify which method to use would be highly advantageous!