# What is a quadratic equation?

A **quadratic equation** is an equation that can be rearranged as a standard form as **ax ^{2} + bx + c = 0**, here a ≠ 0, x is an unknown number also a, b & c are known numbers

If a = 0, then standard equation become **bx + c = 0**, this is a linear equation with solution **x = -c/b** not a quadratic equation.

We can also represent a quadratic equations in a factored form then,

**ax ^{2} + bx + c = 0 = (x-m)(x-n)**, here

**m & n**are roots of the quadratic equation

## How to solve a Quadratic equation?

A quadratic equation has at least one root and at most two roots.

If **b ^{2} – 4ac = 0** or

**b = 0**then the quadratic equation has only one root, If

**b**&

^{2}– 4ac ≠ 0**b ≠ 0**then quadratic equation has two roots.

If b** ^{2}** – 4ac > 0 then quadratic equation has only real roots, If b

**– 4ac < 0 then quadratic equation has only complex roots**

^{2}## 1) Solving quadratic equation by completing the square

We know a quadratic equations is in the form of ax^{2} + bx + c = 0. We can transform this equation in to** (x + u) ^{2} = v **

Let consider, **ax ^{2} + bx + c = 0**

**ax ^{2} + bx = -c**. Divided both sides with

**a**then,

**x ^{2} + bx/a = -c/a**, now add square of the half of the coefficient of

**x**, that is

**(b/2a)**

^{2}**x ^{2} + bx/a + (b/2a)^{2} = -c**

**+**

**.**

**(b/2a)**^{2}We know **(r + s) ^{2} = r^{2} + 2rs + s^{2}**

so, **(x + b/2a) ^{2} = -c + (b/2a)^{2}**. Let

**u = b/2a**and

**v = -c + (b/2a)**then

^{2}**(x + u) ^{2} = v**

take square roots

**x + u = ±√v**

⇒ **x = -u ± √v**

### Example problems using completing the square

##### Q. solve the equation **x**^{2} + 5x + 6 = 0

^{2}+ 5x + 6 = 0

x^{2} + 5x = -6

⇒ x^{2} + 5x + (5/2)^{2}= -6 + (5/2)^{2}

then, (x + 5/2)^{2}= -6 + 25/4 = ¼

x + 5/2 = ±½

so, x = ±½ – 5/2

⇒ **x = -2** and **x = -3**

## 2) Solving quadratic equation by factorisation

We Know a quadratic equations can be represented as a factorised form that is

**ax ^{2} + bx + c = 0 = (x-m)(x-n)** =

**x**here

^{2}– (m+n)x + mn**– (m+n) = b**and

**c = mn**

We guess two numbers **(m & n)** that satisfy the equation for** a** and **b** then these two numbers are the roots of the quadratic equation

### Example problems using factorisation

##### Q. solve the equation **x**^{2} + 5x + 6 = 0

^{2}+ 5x + 6 = 0

From above equation we get **b = -(m + n) = 5** and **c = 6 = mn**

Here we can guess **m = -2** and **n = -3** so solution of the equation are -2 & -3

## 3) Solving with quadratic formula

Quadratic formula is a formula which gives solution of the quadratic equations

We can derive* x* from the quadratic equations that is,

**ax ^{2} + bx + c = 0**

⇒ **ax ^{2} + bx = – c**

so, **x ^{2} + bx/a = – c/a**

⇒ **x ^{2} + bx/a + (b/2a)^{2} = – c/a+ (b/2a)^{2}**

thus, **(x + b/2a)^{2} = – 4ac/4a^{2}+ b^{2}/4a^{2}** =

**(b**

^{2}– 4ac)/4a^{2}**x + b/2a = ±√(b ^{2} – 4ac)/2a**

**x = (-b ±√(b ^{2} – 4ac))/2a**

### Example problems using factorisation

##### Q. solve the equation **x**^{2} + 5x + 6 = 0

^{2}+ 5x + 6 = 0

x = (-5 ±√(5^{2} – 4×1×6))/2×1** **

⇒ x = (-5 ±√(25 – 24))/2** **

⇒ x = (-5 ± 1)/2

then, **x = -2 **&** x = -3**

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