In this chapter we will learn about the concept of quadratic equations with solved examples.

At the end of the chapter, we do also do an exercise of identifying quadratic equations.

## What are quadratic equations ?

The general form of quadratic equation is given as;

\mathtt{ax^{2} +bx+c=\ 0}

Where, “a” cannot be equal to zero.

The word quadratic is made of term ” Quad “ which means square. Hence, quadratic equation always contain variable with power 2.

### Difference between linear equation and quadratic equation ?

In linear equation, the expression contain variables with highest power 1.

The graph of linear equation is always a straight line.

Example of linear equation;
4x + 5 = 0
15 – 7x = 0

Whereas in quadratic equation, the highest power of variable is 2.

The graph of quadratic equation is in form of parabola.

Examples of quadratic equation;

\mathtt{5x^{2} +2x\ +\ 4\ =\ 0}\\\ \\ \mathtt{11x^{2} -7=0}

### Graph of quadratic equation

As mentioned earlier the graph of quadratic equation is in the form of parabola.

Given below are examples of quadratic equation with the respective graph.

(a) \mathtt{x^{2} -4x\ +\ 2\ =\ 0}

(b) \mathtt{x^{2} -9\ =\ 0}

(c) \mathtt{2x^{2} +4x+7\ =\ 0}

## Solving quadratic equation

Consider the following quadratic equation.

\mathtt{x^{2} -8x+4\ =\ 0}

The solution of quadratic equation is;

(a) value of variable x for which whole equation is zero.

(b) the point at graph at which y = 0

In the above graph, A & B are two points where the curve intersect at x axis.

This is the value of x variable where value of y = 0.

Hence, on solving quadratic equation, we are trying to figure the location of these two points.

### Determinants of quadratic equation

Consider the general form of quadratic equation;

\mathtt{ax^{2} +bx+c=\ 0}

Determinant of quadratic equation is given by following formula;

\mathtt{D\ =\ b^{2} -4ac}

If D > 0; we will get real values in the solution.

The graph of quadratic equation with D > 0 is given as;

The curve is intersecting the x axis at two points A & B. The location of the two points is the solution of the quadratic equation.

If D < 0; we will get the imaginary solution.

In this case, the curve will not intersect the x axis. Hence, there is no real solution for this particular case.

I hope you understood the basic concept of quadratic equation. Let us solve some problems for further understanding.

## Identify Quadratic equation – Solved Problems

(01) Identify if the given equations are quadratic equations or not.

\mathtt{( a) \ 5x\ +\ 3\ =\ 0}\\\ \\ \mathtt{( b) \ 2x^{2} +7x\ +\ 9\ =\ 0}\\\ \\ \mathtt{( c) \ x^{2} -25\ =\ 0}\\\ \\ \mathtt{( d) \ x\ +\frac{1}{x} -3\ =\ 0}\\\ \\ \mathtt{( e) \ x^{3} +4x+9\ =\ 0}\\\ \\ \mathtt{( f) \ x^{2} +\frac{2}{x} -3=0}\\\ \\ \mathtt{( g) \ 9x^{2} +2\sqrt{x} -11=\ 0}\\\ \\ \mathtt{( h) \ x^{2} +\ 8\ =\ 0}

Solution

(a) 5x + 3 = 0

Highest power of x is 1.

It is not a quadratic equation.

(b) \mathtt{2x^{2} +7x\ +\ 9\ =\ 0}

Highest power of x is 2.

It is a quadratic equation.

(c) \mathtt{x^{2} -25\ =\ 0}

Highest power of x is 2.

It is a quadratic equation.

(d) \mathtt{x\ +\frac{1}{x} -3\ =\ 0}

Solving the equation.

\mathtt{\frac{x^{2} +1}{x} =3}\\\ \\ \mathtt{x^{2} +1=3x}\\\ \\ \mathtt{x^{2} -3x+1=0}

Note that highest power of x is 2. Hence the expression is quadratic equation.

(e) \mathtt{ \ x^{3} +4x+9\ =\ 0}

Highest power of x is 3.

It’s not a quadratic equation.

(f) \mathtt{x^{2} +\frac{2}{x} -3=0}

Solving the expression further;

\mathtt{\frac{x^{3} +2}{x} \ =\ 3}\\\ \\ \mathtt{x^{3} +2=3x}\\\ \\ \mathtt{x^{3} -3x+2=0}

Highest power of x is 3.

Hence, it is not a quadratic equation.

(g) \mathtt{9x^{2} +2\sqrt{x} -11=\ 0}

Here the power of x is not an integer value.

Hence, it’s not a quadratic equation.

(h) \mathtt{x^{2} +\ 8\ =\ 0}

Highest power of x is 2.

It is a quadratic equation.

(02) Check if the below quadratic equations have real or imaginary solution.

\mathtt{( a) \ x^{2} +5x+\ 6\ =\ 0}\\\ \\ \mathtt{( b) \ x^{2} +4x+5=0}\\\ \\ \mathtt{( c) \ x^{2} +x-2=0}\\\ \\ \mathtt{( d) \ 3x^{2} +x+4}\\\ \\ \mathtt{( e) \ x^{2} +2x^{3} +4x\ =\ 0}

Solution
For each quadratic equation, we have to find the value of determinant.

If D > 0, the solutions are real.

If D < 0, the solutions are imaginary.

Let us solve each equations one by one.

(a) \mathtt{x^{2} +5x+\ 6\ =\ 0}

We know that;

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{\Longrightarrow \ 5^{2} -4( 1)( 6)}\\\ \\ \mathtt{\Longrightarrow \ 25\ -\ 24}\\\ \\ \mathtt{\Longrightarrow \ 1}

Since D > 0, the quadratic equations have real solution.

(b) \mathtt{x^{2} +4x+5=0}

Finding the determinant.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{\Longrightarrow \ 4^{2} -4( 1)( 5)}\\\ \\ \mathtt{\Longrightarrow \ 16\ -\ 20}\\\ \\ \mathtt{\Longrightarrow \ -4}

Since D < 0, the equations have imaginary solution.

(c) \mathtt{x^{2} +x-2=0}

Calculating the determinant.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{\Longrightarrow \ 1^{2} -4( 1)( -2)}\\\ \\ \mathtt{\Longrightarrow \ 1\ +8}\\\ \\ \mathtt{\Longrightarrow \ 9}

Since D > 0, the equations have real solutions.

(d) \mathtt{3x^{2} +x+4}

Calculating the determinant.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{\Longrightarrow \ 1^{2} -4( 3)( 4)}\\\ \\ \mathtt{\Longrightarrow \ 1\ -48}\\\ \\ \mathtt{\Longrightarrow \ -47}

Since D < 0, the equations have imaginary solutions.

(e) \mathtt{\ x^{2} +2x^{3} +4x\ =\ 0}

Highest power of variable is 3.

The expression is not a quadratic equation.

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