Quadratic Equations: Definition, Formula, Solved Examples

A quadratic equation is a second-order equation written as ax² + bx + c = 0 where a, b, and c are coefficients of real numbers and a ≠ 0. Quadratic Equations are used in real-world applications. For example, if school management decides to construct a prayer hall having a carpet area of \(400\) square meters with its length two-meter more than twice its breadth then to find the length and breadth we need the help of a quadratic equation. In this article, let’s discuss Quadratic Equations and their applications in detail.

What are Quadratic Equations?

A polynomial equation of degree \(2\), is called a quadratic equation. 

A quadratic equation in the variable \(x\) is an equation of the form \(a{x^2} + bx + c = 0\), where \({\rm{a, b, c}}\) are real numbers, \(a \ne 0.\)

General Form of a Quadratic Equation

The general form of a Quadratic Equation is given by \(a{x^2} + bx + c = 0\), where \(a,{\rm{ }}b,{\rm{ }}c\) are real numbers, \(a \ne 0.\)

Standard Form of Quadratic Equation

Any equation of the form \(p\left( x \right) = 0\), where \(p\left( x \right)\) is a polynomial of degree\(2\), is a quadratic equation. If we write the terms of \(p\left( x \right)\) in decreasing order of their degrees, then we get the standard form of the equation. That is, \(a{x^2} + bx + c = 0\), \(a \ne 0.\) is called the standard form of a quadratic equation.

Quadratic Equation Examples

Example-1:  \({x^2} – 6x + 8 = 0\)

It is of the form \(a{x^2} + bx + c = 0\). Therefore, it is a quadratic equation.

Example-2: \(x + 12 = 0\) \(x{\rm{ }} + {\rm{ }}12{\rm{ }} = 0\).

It is not of the form \(a{x^2} + bx + c = 0\). Therefore, it is not a quadratic equation.

Example-3:  \({x^2} + 2x + 2 = 0\)

It is of the form \(a{x^2} + bx + c = 0\). Therefore, it is a quadratic equation.

What are the Roots of Quadratic Equation?

In general, a real number \(\alpha \) is called a root of the quadratic equation \(a{x^2} + bx + c = 0,a \ne 0\).  If \(a{\alpha ^2} + b\alpha  + c = 0\) We can say that \(x{\rm{ }} = {\rm{ }}\alpha \)  is a solution 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.

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.

What is Discriminant?

The value of \(\left( {{b^2} – 4ac} \right)\) in the quadratic equation \(a{x^2} + bx + c = 0\). \(a \ne 0\) is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation indicates the nature of roots.

Nature of the Roots of Quadratic Equations 

Since \(\left( {{b^2} – 4ac} \right)\) determines whether the quadratic equation \(a{x^2} + bx + c = 0\) has real roots or not, \(\left( {{b^2} – 4ac} \right)\) is called the discriminant of this quadratic equation. 

So, a quadratic equation \(a{x^2} + bx + c = 0\) has 

1. Two distinct real roots, if \({b^2} – 4ac >0\)
2. Two equal real roots, if \({b^2} – 4ac = 0\)
3. No real roots, if \({b^2} – 4ac < 0\)

Quadratic Equation Formula

If \({b^2} – 4ac \ge 0\), then the roots of the Quadratic Equation \(a{x^2} + bx + c = 0\) are given by \(\frac{ { – b \pm \sqrt { {b^2}-4ac} }}{ {2a}}\). This is the formula for finding the roots of a quadratic equation and it is known as the formula for finding roots of a quadratic equation.

Solving Quadratic Equation By Factorization Method

If we can factorize \(\alpha {x^2} + bx + c,a \ne 0\) , 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.

Example: Find the roots of the quadratic equation \(6{x^2} – x – 2 = 0\)

Solution: We have \(6{x^2} – x – 2 = 0\)

\( \Rightarrow 6{x^2} + 3x – 4x – 2 = 0\)
\( \Rightarrow 3x(2x + 1) – 2(2x + 1) = 0\)
\( \Rightarrow (3x – 2)(2x + 1) = 0\)

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 of are \(\frac23\;\&\;\frac{-1}2\)

Solving Quadratic Equation by Completing The Square Method:

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

Example: \(2{x^2} + 8x + 3 = 0\)

\( \Rightarrow 2{x^2} + 8x =  – 3\) [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\({{{(a + b)}^2} = {a^2} + 2ab + {b^2}}\)]

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 solution.

Solving Quadratic Equation by Formula Method:

If \({b^2} – 4ac \ge 0\) , then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by \(\frac{ { – b \pm \sqrt { {b^2}-4ac} }}{ {2a}}\)

Example: \(3{x^2} – 5x + 2 = 0\)

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

Quadratic Equation formula is given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)

\( = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2 \times 3}} = \frac{{ + 5 \pm \sqrt {25 – 24} }}{6}\)

\( = \frac{{5 \pm \sqrt 1 }}{6} = \frac{{5 \pm 1}}{6} = \frac{{5 + 1}}{6},\frac{{5 – 1}}{6} = \frac{6}{6},\frac{4}{6}\)

\( \Rightarrow x = 1{\;\rm{ or }}\;x = \frac{2}{3}.\)

Hence, the roots of the given quadratic equation are \(1\;\&\;\frac23\)

Graphical Solution of Quadratic Equation

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

\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} – \left( {\frac{D}{{4{a^2}}}} \right)} \right] = 0\)

The above quadratic equation represents a parabola which has vertex at \(P\left[ {\frac{{ – b}}{{2a}},\frac{{ – D}}{{4a}}} \right]\) and axis parallel to \(y\)-axis.

The graphical solution of the above quadratic equation is the two points \(\alpha \) and \(\beta \), where the parabolic graph intersecting the \(x\)-axis as shown below.

Solved Examples – Quadratic Equations:

Question-1: Find the discriminant of the quadratic equation \(2{x^2} – 4x + 3 = 0\) and hence find the nature of its roots.
Answer: The given equation is of the form \(a{x^2} + bx + c = 0\) , 
From the given quadratic equation \({\rm{a = 2}},{\rm{b = }}\,{\rm{ – 4}}\) and \(c = 3.\)
The discriminant \({b^2} – 4ac = {( – 4)^2} – (4 \times 2 \times 3) = 16 – 24 = \, – 8 < 0\)
Therefore, no real roots for the given quadratic equation.

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

Question-3: Find the roots of the quadratic equation \({x^2} + 10x + 21 = 0\) by completing the square method.
Answer: \({x^2} + 10x + 21 = 0\)
\(\Rightarrow {x^2} + 10x = \, – 21\) [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 \(\left. {{{(a + b)}^2} = {a^2} + 2ab + {b^2}} \right]\)
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 \( -3\;\&\;-7 \)

Question-4: Find the discriminant of the quadratic equation \(2{x^2} + 8x + 3 = 0\) and hence find the nature of its roots.
Answer: The given equation is of the form \(a{x^2} + bx + c = 0\), 
From the given quadratic equation \({{a = 2, b = 8}}\) and \(c = 3\).
The discriminant \({b^2} – 4ac = {8^2} – (4 \times 2 \times 3) = 64 – 24 = 40 > 0\)
Therefore, the given quadratic equation has two distinct real roots.

Question-5: Find the roots of the quadratic equation \({x^2} + 3x – 10 = 0\) by factorization method.
Answer: 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 \(2\;\&\;-5\)

Quadratic Equations – Summary

From this article, we learnt that Quadratic Equation is an equation in the form of ax2+bx+c=0.  where a, b, c are real or complex numbers, x is a variable and a ≠ 0. If a =0, then the equation is either linear or quadratic. Any Quadratic Equation can be solved using the factorizing method, completing square method or through quadratic formula.

Frequently Asked Questions (FAQs) – Quadratic Equations

Question-1: What is Discriminant?
Answer: The term \(\left( {{b^2} – 4ac} \right)\) in the quadratic formula is known as the discriminant of a quadratic equation \(a {x^2} + bx + c = 0,a \ne 0\). The discriminant of a quadratic equation shows the nature roots.

Question-2: What are the \(5\) real-life examples of quadratic equation?
Answer: 5 real-life examples, where quadratic equations can be used are (i) Throwing a ball (ii) A parabolic mirror (iii) Shooting a cannon (iv) Diving from a platform and (v) Hitting a golf ball  

Question-3: How do I solve quadratic equations?
Answer: 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 

Question-4: What is the standard form of the quadratic equation?
Answer: The form \(a{x^2} + bx + c = 0,a \ne 0\) is called the standard form of a quadratic equation. 

Question-5: What are the four ways to solve a quadratic equation?
Answer: There are four methods of solving a quadratic equation are
(i) Solving by grouping of the terms and factorizing method
(ii) Completing the square method
(iii) Solving by quadratic formula method
(iv) Graphical method.

Question-6: How can you tell if it is a quadratic equation?
Answer: If a quadratic equation in the variable \(x\) is an equation of the form \(a{x^2} + bx + c = 0\), where \(a,b,c\) are real numbers, \(a \ne 0\).

Now you are provided with all the necessary information about Quadratic Equations. We hope this detailed article is helpful to you. If you have any queries about this article or in general about Quadratic equations, let us know in the comment box below and we will get back to you as soon as possible.