Kiến thức

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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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)\$

#### 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 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}\$

\$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

Problems using Vieta’s formulas

Solution of cubic and quartic equations – 1

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