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.**

**Next chapter** :** Solving quadratic equation using factorization method**