Quadratic Equations
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a = 2, b = 3, c = -5D = (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
Parabola
The graph of a quadratic equatin is called a parabola.
If a > 0 then graph horns pointing down:
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:
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
Quadratic equations problems
Problems using Vieta's formulas
Solution of cubic and quartic equations - 1
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