# What is a quadratic equation?

quadratic equation  is an equation that can be rearranged as a standard form as ax2 + 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,

ax2 + 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 b2 – 4ac = 0 or b = 0 then the quadratic equation has only one root, If b2 – 4ac ≠  0 & b ≠ 0 then quadratic equation has two roots.

If b2 – 4ac > 0 then quadratic equation has only real roots, If b2 – 4ac <  0 then quadratic equation has only complex roots

## 1) Solving quadratic equation by completing the square

We know a quadratic equations is in the form of ax2 + bx + c = 0. We can transform this equation in to (x + u)2 = v

Let consider, ax2 + bx + c = 0

ax2 + bx = -c. Divided both sides with a then,

x2 + bx/a = -c/a, now add square of the half of the coefficient of x, that is (b/2a)2

x2 + bx/a + (b/2a)2 = -c + (b/2a)2.

We know (r + s)2 = r2 + 2rs + s2

so, (x + b/2a)2 = -c + (b/2a)2. Let u = b/2a and v = -c + (b/2a)2 then

(x + u)2 = v

take square roots

x + u = ±√v

x = -u ± √v

### Example problems using completing the square

##### Q. solve the equation x2 + 5x + 6 = 0

x2 + 5x = -6

⇒ x2 + 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

ax2 + bx + c = 0 = (x-m)(x-n) = x2 – (m+n)x + mn here – (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 x2 + 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,

ax2 + bx + c = 0

ax2 + bx = – c

so, x2 + bx/a = – c/a

x2 + bx/a + (b/2a)2 = – c/a+ (b/2a)2

thus, (x + b/2a)2 = – 4ac/4a2+ b2/4a2 = (b2 – 4ac)/4a2

x + b/2a = ±√(b2 – 4ac)/2a

x = (-b ±√(b2 – 4ac))/2a

### Example problems using factorisation

##### Q. solve the equation x2 + 5x + 6 = 0

x = (-5 ±√(52 – 4×1×6))/2×1

⇒ x = (-5 ±√(25 – 24))/2

⇒ x = (-5 ± 1)/2

then, x = -2 & x = -3

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