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

$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The number D = b2 - 4ac is called "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 $x = - \frac{b}{2a}$
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
$x=\frac{-(3) \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} =$
$= \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} =$
$\frac{-3 \pm 5}{2} = \begin{cases} \frac{-3 - 5}{2} = -4 \\ \frac{-3 + 5}{2} = 1\end{cases}$

Standard Form Vertex Form Intercept Form
Function $y=ax^2+bx+c$ $y=a(x-h)^2+k$ $y=a(x-p)(x-q)$
Vertex $\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right)$ $(h, k)$ $\left( \frac{p+q}{2}, f\left(\frac{p+q}{2}\right) \right)$
Axis of Symetry $x=-\frac{b}{2a}$ $x=h$ $x=\frac{p+q}{2}$
Example $y=2x^2+4x-30$ $y=2(x+1)^2-32$ $y=2(x-3)(x+5)$
y-intercept at
$(0, -30)$
vertex at
$(-1, -32)$
x-intercepts at
$(3, 0)$ and $(-5, 0)$

#### Parabola

The graph of a quadratic equation is called a parabola.
If a > 0, then its vertex points down:

If a < 0, then its vertex points up:
If a = 0 the graph is not a parabola and a straight line.

The vertex of the parabola is $x = -\frac{b}{2a}$.

#### Vieta's formulas

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

Problem 1. Solve the equation:
x2 - 4 = 0
Solution: x2 - 4 = (x - 2)(x + 2)
(x - 2)(x + 2) = 0
x - 2 = 0 or x + 2 = 0
The roots are x = 2 or x = -2

Solution 2: a = 1, b = 0, c = -4
D = 02 - 4 ⋅ 1 ⋅ (-4) = 16
$x_1 = \frac{-b - \sqrt{D}}{2a} = \frac{- 0 - \sqrt{16}}{2 \cdot 1} = \frac{-4}{2} = -2$
$x_2 = \frac{-b + \sqrt{D}}{2a} = \frac{- 0 + \sqrt{16}}{2 \cdot 1} = \frac{4}{2} = 2$

Problem 2. Solve the equation:
3x2 + 4x + 5 = 0
Solution: the discriminant is D = 42 - 4⋅3⋅5 = 16 - 60 = -44 < 0
So the quadratic equation has no real roots.

Problem 3. Solve the equation:
x2 + 4x - 5 = 0; x = ?
Solution: The discriminant is 42 - (-4⋅1⋅5) = 16 + 20 = 36 > 0
The equation has 2 real roots: $\frac{-4 \pm \sqrt{36}}{2}$
x = 1 or x = -5

Problem 4. Solve the equation:
x2 + 4x + 4 = 0; x = ?
Solution: The discriminant is 42 - (4⋅1⋅4) = 16 - 16 = 0
So there is one real solution: $x = \frac{-4}{2}$
x = -2

Problem 5. Solve the equation:
x2 - 13x + 12 = 0
Roots: 1, 12

Problem 6. Solve the equation:
8x2 - 30x + 7 = 0
Roots: 3.5, 0.25

x2 x = 0
a = 1, b = 1, c = 1
D = (1)2 - 4⋅1⋅1 = -3
The equation has no real solutions.

#### Derivation of the quadratic formula

By using the "completing the square" technique

$ax^2 + bx + c = 0$
$ax^2 + bx = -c$
$x^2 + \frac{b}{a}x = -\frac{c}{a}$
Add the both sides $\frac{b^2}{4a^2}$

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}= -\frac{c}{a}+ \frac{b^2}{4a^2}$

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}= -\frac{4ac}{4a^2}+ \frac{b^2}{4a^2}$

$(x + \frac{b}{2a})^2= \frac{b^2 - 4ac}{4a^2}$

$x + \frac{b}{2a}= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x = -\frac{b}{2a}\pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

#### Quadratic equations in our math forum

Feedback   Contact email: