An equation of the general form,0

{\color{Purple} ax^{2}+bx+c=0}

Where a, b, and c are constant such that a≠ 0, is called a quadratic equation. In such an equation, the unknown variable “x ” has two as the highest power. Any power above or below two will make the equation polynomial or linear equations respectively.

In this article, we will explain the concept of quadratic equations, explain the various methods of solving quadratic equations with examples and teach you how to find quadratic equations from given roots.

Quadratic equations are usually solved by the following methods:

  1. Factorization
  2. Completing the squares
  3. Formula (By the use of formula) and
  4. Graph

This article will discuss the first three methods, while the subsequent article will discuss the graphical method of solving quadratic equations.

Since the highest power of the unknown is 2, that means there will be two values of the unknown satisfying the equation. These values are sometimes called the ‘roots of the equations‘.

NOTE: If the highest power is 3, there will be 3 roots. If the power is 4, there will be 4 roots and so on.


Consider the general quadratic equation ax^{2}+bx+c=0 To solve this equation by factorization requires factorizing the expression as two linear factors (x-m) and (x+n) such that ax^{2}+bx+c\equiv (x+m)(x+n)

and then using the fact that if (x+m)(x+n)=0, then either

(x+m)=0 ....... (1)  or

(x+n)=0 ....... (2)

resulting in two linear equations 1 and 2 whose solutions form the solutions of the quadratic equations.

NOTE: The roots m and n must be two values such that their product gives +c and their sum is +bx.

Let us consider, some examples of how these two values are calculated.

Example 1: Solve the quadratic equation {\color{Purple} x^{2}+3x-10=0} using the factorization method

Firstly, look for two values whose product/multiplication gives  (−10) and their sum (+3). Those two values can be 5 and 2. Putting the signs, we will have +5 and −2. This is because,

Product: +5 × −2 = −10 and

Sum: +5+(−2) = +5−2 = +3.

To replace +3x in the equation x^{2}+3x-10=0 with +5x and -2x

x^{2}+5x-2x-10=0, simplify further


Since (x-5) appears twice, we pick one part, so

x-2=0 or x+5=0

What if you are giving a question like the one in sample 2, how will you go about it? Now let’s learn how to solve it.

Example 2: Solve using factorization method {\color{Purple} 3x^{2}+5x-2=0}.

From the above question, finding the product and the sum of the value 2, is not possible. What we have to do is multiply 2 by 3, the coefficient of x^{2}, that will give us -6.

So look for two values whose product/multiplication gives  (−6) and their sum (+5). Those two values can be 6 and 1. Putting the signs, we will have +6 and −1. This is because, for the

Product: +6 × −1 = −6 and

Sum: +6+(−1) = +6−1 = +5.

So replace +5x in the equation, 3x^{x}+5x-2=0 with +6x-x (remember x is still the same as 1x).

3x^{2}+6x-x-2=0, simplify further,


So 3x-1=0 and x+2=0

3x=1 and x=-2

x=\frac{1}{3}  or x=-2

Let’s finally consider the last example on factorization method

Example 3: Solve the quadratic equation {\color{Purple} (x-2)^{2}}=9.

To solve this question, the first thing you have to do is to clear the square. To do that we have to square root both sides.

\sqrt{(x-2)}^{2}=\pm \sqrt{9}

From the left hand side, square will clear square root, and for the right hand side, the square root of 9 is 3.

x-2=\pm 3

So x=2+3 or x=2-3

x=5 or x=-1

To check if your result is correct, put x=5 or x=-1 into the equation (x-2)^{2} to see if the result will be 9.

(x-2)^{2}= (5-2)^{2}=3^{2}=9 and (x-2)^{2}=(-1-2)^{2}=(-3)^{2}=9. You See! it’s correct.

NOTE: It is necessary to try to solve a quadratic equation first by factorization method, if it is possible before using any other method, since it is the easiest and quickiest method. 

Let us now consider the other method of solving quadratic equation, the ‘completing the square method’.


Some quadratic equations cannot be solved by factorization method. In such cases, you can use the method of completing the square method. This method involves making the expression of the equation a perfect square which can be factorized.

Consider the equation 2x^{2}+3x-2=0. We are going to use this equation to explain the easy steps involved in completing the square of an equation. Consider the following steps:

STEP 1: Divide through by the coefficient of x^{2} i.e 2 from the above equation. Note, if the coefficient of x is 1, then you should omit this step.



STEP 2: Move the constant term to the right hand side (RHS) (i.e 1 in the above equation).  Remember, that signs changes as it goes across the equality sign.


STEP 3: Make the left hand side (LHS) a perfect square by adding the square of ‘half’ the coefficient of {\color{Purple} x} to both sides of the equations.

The coefficient of x is +\frac{3}{2} . So half of that value is \frac{3}{2} ÷ 2=\frac{3}{4} . The square of the half is (\frac{3}{4})^{2}

x^{2}+ (\frac{3}{4})^{2}=1+(\frac{3}{4})^{2}

That is, (x+\frac{3}{4})^{2}=1+\frac{9}{16}

STEP 4: Factorize the LHS


So (x+\frac{3}{4})^{2}=\frac{25}{16}.

STEP 5: Take the square root of both sides, to clear the square on the RHS. We have 

\sqrt{(x+\frac{3}{4})^{2}}=\pm \sqrt{\frac{25}{16}}

x+\frac{3}{4}=\pm \frac{5}{4}

x=-\frac{3}{4}\pm \frac{5}{4}

x=-\frac{3}{4}+\frac{5}{4} or -\frac{3}{4}-\frac{5}{4}

x=\frac{2}{4}  or -\frac{8}{4}

x=\frac{1}{2} or -2

Let us look at a video example of solving quadratic eqution using the method.

Example 2: Using completing the square method, solve correct to two decimal places  \frac{x-2}{4}=\frac{x+2}{2x}

So you see, solving quadratic equation using completing the square method can be very easy. Let’s consider another example, this time around we will do it straight, knowing very well that you have now master the steps above.

Example 3: Solve the equation 3x^{2}-5x-7=0.

Remember the first step? Let me remind you. Divide through by the coefficient of x^{2} which is 3.

\frac{3}{3}x^{2}-\frac{5}{3}x-\frac{7}{3}=0. This will give us x^{2}-\frac{5}{3}x-\frac{7}{3}=0

Step 2: x^{2}-\frac{5}{3}x=\frac{7}{3}

Step 3: Half of the coefficient of x is -\frac{5}{3} ÷ 2=-\frac{5}{6}. The square of the result is (-\frac{5}{6})^{2}. So add it to both sides to make the RHS a perfect square.




So (x-\frac{5}{6})^{2}=\frac{109}{36}, Square root both sides to clear the square.

\sqrt{(x-\frac{5}{6})}^{2}=\pm \sqrt{\frac{109}{36}}

x-\frac{5}{6}=\pm \sqrt{\frac{109}{36}}

x=\frac{5}{6}+\frac{\sqrt{109}}{6} or \frac{5}{6}-\frac{\sqrt{109}}{6}

x=\frac{5+10.44}{6} or \frac{5-10.44}{6}

x=2.57 or -0.91

We have been able to see from the above example that the method of ‘completing the square‘ method can be used to solve a quadratic equation which cannot be solved by factorization. So the completing the square method is a more general method than the factorization method.

Let us finally consider the last method of solving quadratic equation, which is the quadratic formula method.


Consider the general quadratic equation {\color{Purple} ax^{2}+bx+c=0}

Where a, b, c are constants such that a≠ 0. The roots x of the equation is defined by the formula,

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

Called the quadratic formula. This formula is sometines called the “almighty formula” because it can be used to solve ‘any’ quadratic equation. The quadratic formula can be used when a quadratic equation cannot be solved by factorization or any of the other methods. It is the most general method and can be derived through the method of completing the square as follows:

Given the quadratic equation, ax^{2}+bx+c=0

Divide through by a, to make the coefficient  of x^{2}=1.

ax^{2}+bx+c=0 will become \frac{a}{a}x^{2}+\frac{b}{a}x+\frac{c}{a}=0

Transfer the constant term to the RHS


Make the LHS of the equation a perfect square,

x^{2}+ (\frac{b}{2a})^{2}=-\frac{c}{a}+(\frac{b}{2a})^{2}


Factorize the left hand side (LHS) of the equation,


The left hand side will be \frac{-4ac+b^{2}}{4a^{2}}=\frac{-4ac+b^{2}}{4a^{2}}

So (x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}

Take the square root of both sides:

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

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

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

x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} which is the quadratic formula.

Note: In as much as the quadratic formula solves all quadratic equations, care should be taken in writing the formula correctly, substituting and evaluating correctly in order to aviod mistakes. 

Example 1: Solve the equation {\color{Purple} 2x^{2}+15x+7=0} using the quadratic formula.

Compare the equation 2x^{2}+15x+7=0 with ax^{2}+bx+c=0. We will see that a=2, b=15 and c=7.

Substitue these values into the formula, x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} we will get

x=\frac{-15\pm \sqrt{15^{2}-4\star 2\star 7}}{2\star 2}

x=\frac{-15\pm \sqrt{225-54}}{4}

x=\frac{-15\pm \sqrt{169}}{4}\frac{-15\pm 13}{4}

This will result to  \frac{-15+13}{4} or \frac{-15-13}{4}

\therefore x= -\frac{1}{2} or -\frac{28}{4}

=-\frac{1}{2} or -7

Note: If giving the equation ax^{2}+bx+c=0 as a quadratic equation to solve, the following results should be the possible outcome:

  • If b^{2}-4ac> 0 then the roots are real and district;
  • If b^{2}-4ac< 0, then the roots are imaginary or complex and
  • If b^{2}=4ac=0, then the roots are real and equal.

Having concluded our discussions on the various methods of solving quadratic equation, there is one finally thing I want you to learn, that is the formulation of quadratic equations from roots. The use to give a lot of students issues. But our discussion will booast your confidence in solving such questions in examinations.


An equation can be found it its roots are given. So given that the roots of a quadratic equation are m and n, then either x=m or x=n. That is, x-m=0 or x-n=0.

So (x-m)(x-n)=0

\therefore x^{2}-mx-nx+mn=0

That is x^{2}-(m+n)x+mn=0

This is the required equation which has the following properties:

  1. The cofficient of x^{2} is unity;
  2. The sum of the roots m+n is the coefficient of x with the sign changed;
  3. The product of the roots mn is the constant term.

Thus, if the roots of the quadratic equation are given, the the equation can be obtained (with the usual three terms) as

x^{2}- (sum of roots)x + (product of roots)=0

We shall now illustrate this fact with two examples.

Example 1: Find the quadratic equation whose roots are 5 and 6.

Let m=5 and n=6

\therefore m+n=5+6=11 and mn=5\times 6=30

Therefore, the quadratic equation whose roots are 5 and 6 is x^{2}-11x+30=0

Or we can go about it this other way I call it the “Bottom – Top Approach”, choose which is easier for you to grasp easily,

Given that x=5 and x=6

Then x-5=0 and x-6=0

So (x-5)(x-6)=0

x^{2}-6x-5x+30=0 when simplified we will get

\therefore x^{2}-11x+30=0

So you see, the results are the same, whichever method is fine and easier for you, use it!

Example 2: Find the quadratic equation whose roots are \frac{1}{2}  and -\frac{3}{4} .

Let α =\frac{1}{2} and β=-\frac{3}{4}

Then α + β=\frac{1}{2}+(-\frac{3}{4})=(-\frac{1}{4}) and αβ=(\frac{1}{2})(-\frac{3}{4})=-\frac{3}{8}

Therefore, the equation is x^{2}-(-\frac{1}{4})x+(-\frac{3}{8})=0

That will be simplified to give x^{2}+\frac{1}{4}x-\frac{3}{8}=0

Multiply through by 8,


In conclusion, solving quadratic equations using the above methods are very easy. Learn the steps, master them, and be careful in applying them to aviod making mistake.

Download a copy of the Quadratic Equation Worksheet for quizzes on this topic.

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September 17, 2021

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