Quadratic Equations

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a = 2, b = 3, c = -5
D = (3)² - 4⋅2⋅(-5) = 49
D = 7²
The solutions are:
x₁ = (-3 + √49)/(2⋅2) = (-3 + 7)/4 = 1
x₂ = (-3 - √49)/(2⋅2) = (-3 - 7)/4 = -2.5

Quadratic equations looks like: ax² + bx + c = 0
where a,b,c are real numbers, and a ≠ 0. Every quadratic equation can have 0, 1 or 2 real decidions derived by the formula:

The number D = b² - 4ac is called the "discriminant".
If D < 0 then the quadratic equation has no real solutions(it has 2 complex solutions).
If D = 0 then the quadratic equation has 1 solution that is given by the formula: x = - ^b/_2a.
If D > 0 then the quadratic equation has 2 distinct solutions.

Example:
Let's solve the quadratic equation: x² + 3x - 4 = 0
a = 1, b = 3, c = -4
example of solving quadratic equation

Parabola

The graph of a quadratic equatin is called a parabola.
If a > 0 then graph horns pointing down:

if a < 0 then graph horns pointing up:

The midpoint of any parabola is the point x = -b/2a.

Vieta's formulas

If x₁ and x₂ are the roots of the quadratic equation ax² + bx + c = 0 then:
$x_1 + x_2 = -\frac{b}{a} \qquad \qquad \qquad x_1x_2 = \frac{c}{a}$
These formulas are called Vieta's formulas.
We can find the roots x₁ and x₂ of a quadratic equation by solving the system above.

Quadratic Equation Problems

1) x² - 4 = 0; x = ?
Solution: x² - 4 = (x - 2)(x + 2)
x = 2 or x = -2

2) 3x² + 4x + 5 = 0; x = ?
Solution: the discriminant is 4² - 4.3.5 = 16 - 60 = -44 < 0 So the quadratic equation have no real decisions.

3) x² + 4x - 5 = 0; x = ?
Solution: The discriminant is 4² - (-4.1.5) = 16 + 20 = 36 > 0 So there are two real solutions: ¹/₂(-4 ± √36)
x = 1 or x = -5

4) x² + 4x + 4 = 0; x = ?
Solution: The discriminant is 4² - (4.1.4) = 16 - 16 = 0 So there is one real solution: ¹/₂(-4)
x = -2

5) x² - 13x + 12 = 0
Solution: 1, 12
Draw the graph of the function: f(x) = x² - 13x + 12

6) 8x² - 30x + 7 = 0
Solution: 3.5, 0.25
Draw the graph of the function: f(x) = 8x² - 30x + 7

Additional quadratic equation problems

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Problems using Vieta's formulas
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