A quadratic is a type of problem In mathematics that deals with a variable multiplied by itself (an operation known as squaring).

It derives from the area of a square being its side length multiplied by itself. The word quadratic comes from “quadratum”, which means square in Latin word.

There are a lot of phenomena in the real world that can be described by Quadratic equations. For example is rocket fly route, where it will land, how long it will take a person to row down & up a river, or how much to charge for a product.

Because of these applications, quadratics equations have profound historical importance and were foundation of algebra.

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## Quadratic Equation

Only if it can be put in the form *ax*^{2}* + bx + c *= 0, and a is not zero.

The name comes from “quad” meaning square, as the variable is squared (in other words x2).

These are all quadratic equations in disguise:

Equation | In standard form | a, b and c |

x^{2} = x + 1 | x² -x – 3 = 0 | a=1, b=-1, c=-3 |

2(x^{2} – 2x) = 2 | 2x² – 4x – 2 = 0 | a=2, b=-4, c=-2 |

x(x-6) = 18 | -x² +6x + 18 = 0 | a=-1, b=6, c=18 |

20 – 15/x – 10/x^{2} = 0 | 20x² -15x – 10 = 0 | a=20, b=-15, c=-10 |

## Quadratic Formula

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula:

The “±” means we need to do a plus AND a minus, so there are normally TWO solutions !

The part **(b2 – 4ac)** is called the “** discriminant**“, because it can “discriminate” between the possible types of answer:

- when b2 – 4ac > 0, we get two real solutions,
- when b2 – 4ac = 0, we get just ONE solution,
- when b2 – 4ac < 0, we get complex solutions.

Find x !

Equation | a, b and c | Operation | x |

x² -x – 3 = 0 | a=1, b=-1, c=-3 | ||

2x² – 4x – 2 = 0 | a=2, b=-4, c=-2 |

In order to Quadratic Formula to work, you must have your equation arranged in the form:

**(quadratic) = 0**

Also, the 2a in the denominator of the Formula is not just the square root. It is underneath everything above.

Make sure not to drop the square root or the “±” in the middle of your calculations, so you will not need to put them back in.

Remember that b^{2} means “the square of ALL of b, including its sign”. Do not leave b^{2} being negative, even if b is negative, because the square of a negative is a positive.

In other words, don’t be sloppy and don’t try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

## Examples

1. Solve the quadratic equation by factorization method:

a. x^{2} – 4x – 12 = 0

b. m^{2} = -6m – 5

a. x^{2} – 4x – 12 = 0

it can be solved by using factorization.

x^{2} – 4x – 12 = 0

→ (x-6)(x+2) = 0

→ x_{1} = 6

→ x_{2} = -2

b. m^{2} = -6m – 5

m^{2} + 6m + 5 = 0 (using factorization)

→ (m+5)(m+1) = 0

→ m_{1} = -5

→m_{2} = -1

2. Solve the quadratic equations:

a. 2x^{2} – 3x -2 = 0

b. k(k+2) + 2 = 0

c. (k-2)^{2 }– 36 = 0

a. 2x^{2} – 3x -2 = 0

x_{1,2} = {-b ± √(b^{2}-4ac)} / 2a

= {3 ± √(9-4.2.(-2))} / 2.2

= {3 ± √25} / 4

= {3 ± 5} / 4

x_{1} = 8/4

x_{2} = -2/4 = – ½

b. k(k+2) + 2 = 0 → k^{2} + 2k +2 = 0

x_{1,2} = {-b ± √(b^{2}-4ac)} / 2a

= {-2 ± √(4-4.1.(2))} / 2.1

= {-2 ± √-4} / 2

= {-2 ± 2√-1} / 2

= {-2 ± 2i}/2

= -1 ± i

x_{1} = -1 + i

x_{2} = -1 – i

c. (k-2)^{2 }– 36 = 0 → k^{2} -4k + 4 -36 = 0 → k^{2} – 4k – 32 = 0

x_{1,2} = {-b ± √(b^{2}-4ac)} / 2a

= {4 ± √(16-4.1.(-32))} / 2.1

= {4 ± √144} / 2

= {4 ± 12} / 2

= 2 ± 6

x_{1} = 8

x_{2} = -4

3. solve the quadratic equation in fraction forms:

a. {(x^{2}-10)/x+2} + x – 1 = x

b. y + 1 = {(2y–7)/(y+5)} – {(5y+8)/(y+5)}

a. {(x^{2}-10)/x+2} + x – 1 = x

→ {(x^{2}-10)/x+2} = 1 + x – x

{(x^{2}-10)/x+2} = 1

{(x^{2}-10)/x+2} (x+2) = 1 (x+2) (simplify by multiplying the denominator)

x^{2 }– 10 -x- 2 = 0

x^{2 }– x- 12 = 0

(x+3)(x-4) = 0

x_{1} = -3

x_{2} = 4

b. y + 1 = {(2y–7)/(y+5)} – {(5y+8)/(y+5)}

y + 1 = {(2y–7-(5y+8))/(y+5)}

y + 1 = {(2y–7-5y-8))/(y+5)}

y + 1 = {(-3y–15)) / (y+5)}

(y + 1)(y+5) = {(-3y–15)) / (y+5)} (y+5) (simplify by multiplying the denominator)

y^{2} + 6y + 5 = -3y – 15

y^{2} + 9y + 20 = 0

(y+4)(y+5) = 0

x_{1} = -4

x_{2} = -5

4. Alex has some box of candies. In mathematics form, it is x^{2} – 2x – 3 = 0. x as a box that contain some candies. If he bought a candy $0.5 then how much he did paid if he only bought 4box?

Candies x^{2} – 2x – 3 = 0, x as the box that contain some candies.

First step is determining how many candies in a box. In another words, we must determine the value of x.

x^{2} – 2x – 3 = 0

(x-3)(x+2) = 0

x_{1} = 3

x_{2} = -1

(it is impossible x = -1 because of the content is candy. It is impossible if it minus / negative)

Next, each candy is $0.5. So, if Alex bought 4box means 4x = 4 (3) = 12 candies.

The price is 12 x (0.5) = $6.