# Quadratic Equations

**Quadratic equations** looks like: ax^{2} + 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:

The number D = b^{2} - 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: x^{2} + 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} = $

$\frac{-3 \pm 5}{2} = \begin{cases} \frac{-3 - 5}{2} = -4 \\ \frac{-3 + 5}{2} = 1\end{cases}$

#### Parabola

The graph of a quadratic equation is called a **parabola**.

If a > 0, then its vertex points down:

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

#### Vieta's formulas

If x_{1} and x_{2} are the roots of the quadratic equation ax^{2} + 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 x_{1} and x_{2} of the quadratic equation by solving the simultaneous equations.

#### Problems involving quadratic equations

**Problem 1.** Solve the equation:

x^{2} - 4 = 0

**Solution:** x^{2} - 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 = 0^{2} - 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:

3x^{2} + 4x + 5 = 0

**Solution:** the discriminant is D = 4^{2} - 4⋅3⋅5 = 16 - 60 = -44 < 0

So the quadratic equation has no real roots.

**Problem 3.** Solve the equation:

x^{2} + 4x - 5 = 0; x = ?

**Solution:** The discriminant is 4^{2} - (-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:

x^{2} + 4x + 4 = 0; x = ?

**Solution:** The discriminant is 4^{2} - (4⋅1⋅4) = 16 - 16 = 0

So there is one real solution: $x = \frac{-4}{2}$

x = -2

**Problem 5.** Solve the equation:

x^{2} - 13x + 12 = 0

**Roots:** 1, 12

**Problem 6.** Solve the equation:

8x^{2} - 30x + 7 = 0

**Roots:** 3.5, 0.25

#### Additional resources involving quadratic equations

Quadratic equations problems

Problems using Vieta's formulas

Solution of cubic and quartic equations - 1

#### Free quadratic equation solver

a = 1, b = 1, c = 1D = (1)

^{2}- 4⋅1⋅1 = -3

The equation has no real solutions.