Methods of Solving Quadratic Equations: Formula, Methods, Examples

In algebra, polynomials are algebraic expressions with exponents of the variables as whole numbers. When a polynomial is equated to zero, we get a polynomial equation. If a quadratic polynomial is equated to zero, then we can call it a quadratic equation. A quadratic equation is an equation of second degree.

A quadratic equation makes a \( \cup \)-shaped curve (parabola) if we represent it graphically. There are always two solutions to a quadratic equation with either real roots or imaginary roots. Let us learn in detail the different methods of solving quadratic equations.

Definition of Quadratic Equations

An equation of second-degree polynomial in one variable, such as \(x\) usually equated to zero, is called a quadratic equation. The coefficient of \({x^2}\) must not be zero in a quadratic equation.
\(p(x)\) is a quadratic polynomial, then \(p(x) = 0\) called a quadratic equation.

For example, \(3{x^2} + 2x + 2 = 0, – {x^2} + 6x + 1 = 0,7{x^2} – 6x + 4 = 0\)etc., are quadratic equations.

Standard Form of a Quadratic Equation    

The standard form of a quadratic equation is given by \(a{x^2} + bx + c = 0\) where\(a,b,c\) are real numbers, \(a \ne 0\) and \(a\) is the coefficient of \({x^2},b\) is the coefficient of \(x\) and \(c\) is a constant.

Roots of Quadratic Equations

The values of the variable, like \(x\) that satisfy the equation in one variable are called the roots of the equation. The roots of the quadratic equation may be real or imaginary.

If a quadratic polynomial is equated to zero, it becomes a quadratic equation. The values of \(x\) satisfying the equation are known as the roots of the quadratic equation. Note that the zeroes of the quadratic polynomial \(a{x^2} + bx + c\) and the roots of the quadratic equation \(a{x^2} + bx + c = 0\) are the same.

Methods to Solve the Quadratic Equations

There are some methods to solve the quadratic equation. They are,

1. Factorization method
2. Completing square method
3. Formula method
4. Graphical Method

Solving Quadratic Equation by Factorization Method

If we can factorize \(a{x^2} + bx + c,\,a \ne 0,\) into a product of two linear factors, then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) can be found by equating each factor to zero.

Steps to Solve Quadratic Equation Using Factorization

If we can factorize \(a{x^2} + bx + c = 0,\,a \ne 0,\) into a product of two linear factors, then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) can be found by equating each factor to zero.

Let us take an example and discuss it.

Consider a quadratic equation \({x^2} + 5x + 6 = 0.\)

1. Compare the given quadratic equation with the standard form \(a{x^2} + bx + c = 0\) and find the coefficients of \({x^2},x\) and the constant to get the values for \(a,b,c.\)

So, comparing \({x^2} + 5x + 6 = 0\) with \(a{x^2} + bx + c = 0\) we get, \(a = 1,b = 5,c = 6.\)

2. Now we will split \(b\) as the sum of two numbers such that the product of these two numbers \( = a \times c = ac\)
We can factorize a quadratic equation when \(b\) can be split in \(v\) and \(w\) or \(b = v + w\) and \( = a\, \times c = ac.\)

So, \({x^2} + 5x + 6 = 0 \Rightarrow {x^2} + (2 + 3)x + 6 = 0 \Rightarrow {x^2} + 2x + 3x + 6 = 0.\)

3. Find the common factor of the first two and last two terms separately, such as the new two terms have the same common factor.
So, \(x\) the common factor of the first two terms \({x^2} + 2x.\) Therefore, taking \(x\) outside, we get \({x^2} + 2x = x(x + 2).\)

\(3\) is the common factor of the last two terms \(3x + 6.\) Therefore, taking \(3\) outside, we get \(3x+6=3(x+2).\)

So, \({x^2} + 2x + 3x + 6 = 0\) can be written as \(x(x + 2) + 3(x + 2) = 0.\)

4. If we take out the same common factor, then we get the product of two linear polynomials.
Here, we can see that \((x + 2)\) is the common factor in the new two terms.
So, we can write \(x(x + 2) + 3(x + 2) = 0\,as\,(x + 2)(x + 3) = 0\)

5. At last, we will use the zero product rule to find the zeros. Zero product property says that when \(p \times q = 0\) then either \(p = 0\,or\,q = 0\)
Therefore, \((x + 2) = 0,or(x + 3) = 0.\)

6. Hence, the solutions or roots of the quadratic equation \({x^2} + 5x + 6 = 0\) are \(x =  – 2,x =  – 3.\)

Solving Quadratic Equation by Completing the Square Method

A quadratic equation can be solved by the method of completing the square.

Steps to Solve Quadratic Equation by Completing the Square Method

Consider the quadratic equation, \(a{x^2} + bx + c = 0,a \ne 0\).
Let us divide the equation by \(a\).
\({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)
Multiply and divide \(2\) to \(x\) term.
\( \Rightarrow {x^2} + 2 \times \frac{b}{{2a}} \times x + \frac{c}{a} = 0\)

Now, to make the perfect square, we need to add and subtract \({\left( {\frac{b}{{2a}}} \right)^2}\) from L.H.S.
\( \Rightarrow {x^2} + 2 \times \frac{b}{{2a}} \times x + {\left( {\frac{b}{{2a}}} \right)^2} – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a} = 0\)
\( \Rightarrow {\left( {x + \frac{b}{{2a}}} \right)^2} – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a} = 0\)

Transferring \( – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a}\) from \(L.H.S\) to \(R.H.S\) we have,
\( \Rightarrow {\left( {x + \frac{b}{{2a}}} \right)^2} =  – \frac{c}{a} + {\left( {\frac{b}{{2a}}} \right)^2}\)

Taking square root on both sides we have,
\( \Rightarrow x + \frac{b}{{2a}} =  \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}} \)
\( \Rightarrow x =  \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}}  – \frac{b}{{2a}}\)

For example, consider the quadratic equation \(2{x^2} + 8x + 3 = 0\)

Let us divide the equation by \(a\, = \,2\)
\(2\left( {{x^2} + \frac{8}{2}x + \frac{3}{2}} \right) = 0\)
\( \Rightarrow {x^2} + 4x + \frac{3}{2} = 0\)
\( \Rightarrow {x^2} + 2 \times 2x + \frac{3}{2} = 0\)

Now, to make the perfect square, we need to add and subtract \({(2)^2}\) from \(L.H.S.\)
\( \Rightarrow {x^2} + 2 \times 2 \times x + {(2)^2} – {(2)^2} + \frac{3}{2} = 0\)
\( \Rightarrow {(x + 2)^2} – {(2)^2} + \frac{3}{2} = 0\)

Transferring \( – {(2)^2} + \frac{3}{2}\) from \(L.H.S\,to\,R.H.S\) we have,
\( \Rightarrow {(x + 2)^2} = {(2)^2} – \frac{3}{2}\)

Taking square root on both sides we have,
\( \Rightarrow x + 2 =  \pm \sqrt {4 – \frac{3}{2}} \)
\( \Rightarrow x =  \pm \sqrt {\frac{5}{2}}  – 2\)

Hence, the required solution of the quadratic equation \(2{x^2} + 8x + 3 = 0\) is \(x =  \pm \sqrt {\frac{5}{2}}  – 2\)

Solving Quadratic Equation by Formula Method

The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by the quadratic formula

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

For example, consider the quadratic equation \(3{x^2} – 5x + 2 = 0\)

From the given quadratic equation \(a = 3,\,b =  – 5,\,c = 2\)

The quadratic Equation formula is given by

\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\(x = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2a}} = \frac{{ + 5 \pm \sqrt {25 – 24} }}{6}\)
\( = x = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2 \times 3}} = \frac{{5 \pm \sqrt {25 – 24} }}{6}\)
\( = \frac{{5 \pm \sqrt 1 }}{6} = \frac{{5 \pm 1}}{6} = \frac{{5 + 1}}{6},\frac{{5 – 1}}{6} = \frac{6}{6},\frac{4}{6}\)
\( \Rightarrow x = 1\) or \(x = \frac{2}{3}\)

Hence, the roots of the given quadratic equation are \( \Rightarrow x = 1\) or \(x = \frac{2}{3}\)

Graphical Solution of Quadratic Equation

The standard form of a quadratic equation is \(a{x^2} + bx + c = 0,\) where \(a,b\) and \(c\) are real and \(a\,\, \ne \,\,0\)

Let us take an example and discuss \({x^2} – 5x + 6 = 0\)
Considering \(p(x) = {x^2} – 5x + 6\)
A quadratic polynomial always gives a parabolic graph. In a polynomial, the value of \(x\) which is responsible to make \(p(x) = 0\) is called the zero of the polynomial. When the polynomial equated with zero, it becomes an equation. If the zeros of the quadratic polynomial are known then they can be considered as the solutions of the equation which is formed by equating the polynomial to zero.

When \(x =  + 2,p(2) = {(2)^2} – 5(2) + 6 = 4 – 10 + 6 =  – 6 + 6 = 0\)
\(x =  + 3,p(3) = {(3)^2} – 5(3) + 6 = 9 – 15 + 6 =  – 6 + 6 = 0\)
Hence, the zeros of the quadratic polynomial are \( + 3,\, + 2\)
If we represent the polynomial graphically then, the graph of the polynomial cuts the \(x – \)axis at \( + 3\,{\rm{and}}\, + 2\)

Let us take another example and discuss \({x^2} – 2x + 1 = 0\).
Considering \(p(x) = {x^2} – 2x + 1\)
A quadratic polynomial always gives a parabolic graph.
When \(x =  + 1,p(1) = {(1)^2} – 2(1) + 1 = 1 – 2 + 1 =  – 1 + 1 = 0\)
Hence, the zeros of the polynomial are \( + 1\)
If we represent the polynomial graphically then, the graph of the polynomial touches the \(x – \)axis at \( + 1\).
Therefore, the solution of the equation is \( + 1\).

Solved Examples

Q.1. Find the roots of the quadratic equation \(6{x^2} – x – 2 = 0\)
Ans: We have \(6{x^2} – x – 2 = 0\)
\( \Rightarrow 6{x^2} + 3x – 4x – 2 = 0\)
\( \Rightarrow 3x(2x + 1) – 2(2x + 1) = 0\)
The roots of \(6{x^2} – x – 2 = 0\) are the values of \(x\) for which \((3x – 2)(2x + 1) = 0\)
Therefore, \(3x – 2 = 0\,\,or\,2x\,\, + \,\,1 = 0\)
\(x = \frac{2}{3},x = \frac{{ – 1}}{2}\)
Hence, the roots are \(\frac{2}{3}\& \frac{{ – 1}}{2}.\)

Q.2. Find the roots of the quadratic equation by using the formula method \(2{x^2} – 8x – 24 = 0\)
Ans: From the given quadratic equation \(a = 2,b =  – 8,c =  – 24\)
Quadratic equation formula is given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\(x = \frac{{ – ( – 8) \pm \sqrt {{{( – 8)}^2} – 4 \times 2 \times ( – 24)} }}{{2 \times 2}} = \frac{{8 \pm \sqrt {64 + 192} }}{4}\)
\(x = \,\frac{{8 \pm \sqrt {256} }}{4}S = \frac{{8 \pm 16}}{4} = \frac{{8 + 16}}{4},\frac{{8 – 16}}{4} = \frac{{24}}{4},\frac{{ – 8}}{4}\)
\( \Rightarrow x = 6,x =  – 2\)
Hence, the roots of the given quadratic equation are \(6\&  – 2.\)

Q.3. Find the roots of the quadratic equation \({x^2} + 10x + 21 = 0\) by completing the square method.
Ans: \({x^2} + 10x + 21 = 0\)
\( \Rightarrow {x^2} + 10x =  – 21\) (Subtracted \(21\) from both sides of the equation)
\( \Rightarrow {x^2} + 10x + 25 =  – 21 + 25\) (Added \({\left( {\frac{b}{2}} \right)^2} = {\left( {\frac{{10}}{2}} \right)^2} = 25\) on both the sides of the equation)
\( \Rightarrow {(x + 5)^2} = 4\) (Completed the square by using the identity \({(a + b)^2} = {a^2} + 2ab + {b^2}\))
Then, take the square root on both sides.
\(x + 5 =  \pm 2\)
\(x =  – 3,x =  – 7\)
Hence, the roots of the given quadratic equation are \( – 3\&  – 7.\)

Q.4. Find the roots of the quadratic equation\(2{x^2} + 8x + 3 = 0\) by completing the square method.
Ans: \( \Rightarrow 2{x^2} + 8x =  – 3\) [Subtracted \(3\) from both sides of the equation]
\( \Rightarrow {x^2} + 4x = \frac{{ – 3}}{2}\) (Divided both sides of the equation by \(2\))
\( \Rightarrow {x^2} + 4x+4 = \frac{{ – 3}}{2} + 4\) [Added \({\left( {\frac{b}{2}} \right)^2} = {\left( {\frac{4}{2}} \right)^2} = 4\) on both the sides of the equation]
\(\Rightarrow(x+2)^{2}=\frac{5}{2}\) [Completed the square by using the identity \((a+b)^{2}=a^{2}+2 a b+b^{2}\)]
Then, take the square root on both the sides \( \Rightarrow x + 2 =  \pm \sqrt {\frac{5}{2}} \)
\( \Rightarrow x = -2 \pm \sqrt {\frac{5}{2}} \) is the required solutions.

Q.5. Find the roots of the quadratic equation \({x^2} + 3x – 10 = 0\) by factorization method.
Ans: We have \({x^2} + 3x – 10 = 0\)
\( \Rightarrow {x^2} + 5x – 2x – 10 = 0\)
\( \Rightarrow x(x + 5) – 2(x + 5) = 0\)
\( \Rightarrow (x – 2)(x + 5) = 0\)
So, the roots of \({x^2} + 3x – 10 = 0\) are the values of \(x\) for which \((x – 2)(x + 5) = 0\)
Therefore, \(x – 2 = 0\,\,or\,x + 5 = 0\)
\(x = 2, – 5\,\)
Hence, the roots are \(2\&  – 5.\)

Summary

In this article, we discussed quadratic equation in the variable \(x\) which is an equation of the form \(a{x^2} + bx + c = 0,\) where \(a,b,c\) are real numbers, \(a\, \ne 0\) Also, we discussed the methods of solving the quadratic equations such as factorizing method, completing the square method, formula method etc.

FAQs on Methods of Solving Quadratic Equations

Q.1. What are \(5\) methods of solving a quadratic equation?
Ans: We can solve the quadratic equations by using different methods given below:
1. By factorizing method
2. By completing the square method
3. By using the quadratic formula
4. By using the graphical method
5. By using the trial and error method

Q.2. How can you identify a quadratic equation?
Ans: An equation is a quadratic equation in the variable \(x\) if it is of the form \(a{x^2} + bx + c = 0\) where \(a,b,c\) are real numbers, \(a\, \ne \,0\).

Q.3. Which method can you use to solve all quadratic equations?
Ans: We can not use factorizing method and completing square method for every quadratic equation as there are some constraints. We can use the formula method to solve all quadratic equations.
The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by:
\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
This is the quadratic formula for finding the roots of a quadratic equation.

Q.4. How many types of quadratic equations are there?
Ans: There are three types of quadratic equations:
1. Standard form: \(a{x^2} + bx + c = 0,a \ne 0\)
2. Factored form: \((x – a)(x – b) = 0\)
3. Vertex form: \(a{(x – h)^2} + k = 0\)
Each form of a quadratic equation has specific importance. Therefore, identifying the benefits of each different form can make it easier to understand and solve different situations.

Q.5. Can you use the quadratic formula for any quadratic equation?
Ans: Yes, we can use the quadratic formula for any quadratic equation.

Q.6. What is the standard form of the quadratic equation?
Ans: The form \(a{x^2} + bx + c = 0,\,a \ne 0\) is called the standard form of a quadratic equation.

We hope you find this detailed article on methods of solving quadratic equations helpful. If you have any doubts or queries regarding this topic, feel to ask us in the comment section.