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:
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} = $
$\frac{-3 \pm 5}{2} = \begin{cases} \frac{-3 - 5}{2} = -4 \\ \frac{-3 + 5}{2} = 1\end{cases}$
Quadratic Function
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:
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.
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
Free quadratic equation solver
a = 1, b = 1, c = 1D = (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
Quadratic equations problems
Problems using Vieta's formulas
Solution of cubic and quartic equations - 1