Quadratic Equations

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

Parabola

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

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

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

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.

Problems involving quadratic 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

Additional resources involving quadratic equations

Quadratic equations problems
Problems using Vieta's formulas
Solution of cubic and quartic equations - 1

Free quadratic equation solver

x2 x = 0
a = 2, b = 3, c = -5
D = (3)2 - 4⋅2⋅(-5) = 49
D = 72
The solutions are:
$x_1 = \frac{-3 + \sqrt{49}}{2 \cdot 2} = \frac{-3 + 7}{4} = 1$
$x_2 = \frac{-3 - \sqrt{49}}{2\cdot 2} = \frac{-3 - 7}{4} = -2.5$

Quadratic equations in our math forum

Forums involving quadratic equations


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