Kiến thức

Quadratic Equation, Discriminant, Parabola

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

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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 $left( -frac{b}{2a}, fleft(-frac{b}{2a}right) right)$ $(h, k)$ $left( frac{p+q}{2}, fleft(frac{p+q}{2}right) right)$
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 $(-5, 0)$

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Parabola

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

parabola with vertex downparabola with vertex down

If a < 0, then its vertex points up:

parabola with vertex upparabola with vertex 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.

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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 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

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

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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

Forums involving quadratic equations

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