# EQUATION SOLVER 4in1Examples ho to solve equations

## Linear Equation Solving

Example 1 (Difficulty level - 2):
Let's solve the following linear equation:
x - 4 × (1 - 2x) = 8 - x

Solution:
x - 4 × (1 - 2x) = 8 - x
x - 4 + 2x = 8 - x
3x - 4 = 8 - x
3x + x = 8 + 4
4x = 12
4x = 12 | ÷ 4 x = 3
Answer: x = 3

## Quadratic Equation Solving

Example 2 (Difficulty level - 4):
Let's solve the next quadratic equation:
x2 - 6
2
x + 17 = 0

Solution:
Let's calculate the discriminant of the square trinomial.
We have
a = 1, b = - 6
2
, c = 17

Since
b = - 6
2
it is possible to divide it by two.
So
b/2
= - 3
2

Let's compute the discriminant D1:
D1 = (
b/2
)2 - ac = ( - 3
2
)2 - 1 × 17 = 18 - 17 = 1 > 0

Therefore the equation has two real roots.
x1,2 =
-
b/2
±
D1
/a

x1,2 =
-
3
2
±
1
/1

x1,2 = 3
2
± 1

x1 = 3
2
+ 1

x2 = 3
2
- 1

x1 = 3
2
+ 1, x2 = 3
2
- 1

## Cubic Equation Solving

Example 3 (Difficulty level - 5):
Let's solve the following third degree equation:
3x3 + 4x2 - 4x - 3 = 0

Solution:
Let's factor the left part of original cubic equation, using the cubes difference formula:
a3 – b3 = (a - b)(a2 + ab + b2)

3x3 + 4x2 - 4x - 3 = 3(x3 - 1) + 4x2 - 4x = 3(x - 1)(x2 + x + 1) + 4x(x - 1) = (x - 1)(3x2 + 3x + 3 + 4x) = (x - 1)(3x2 + 7x + 3)

So the original cubic equation is equivalent to the following set of the equations:
x - 1 = 0

3x2 + 7x + 3 = 0

The first equation of the set is a linear equation and has the root x = -1.
The second equation is a quadratic equation solved by the quadratic formula, so the roots of the square equation are:
x1,2 =
- b ±
b2 - 4 × a × c
/2 × a

x1,2 =
- 7 ±
72 - 4 × 3 × 3
/2 × 3

x1,2 =
- 7 ±
49 - 36
/6

x1,2 =
- 7 ±
13
/6

x1 =
- 7 +
13
/6

x2 =
- 7 -
13
/6

x1 = -1, x2 =
- 7 +
13
/6
, x3 =
- 7 -
13
/6

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