# Tag Archives: formula

## Distance formula

Given two points (x 1, y1), (x 2, y2) the formula for distance is calculated with the following formula.

Example # 1:

Use the formula distance to find the distance between (2,3) and (6.6)

Let (x 1, y1) = (2,3)

Let (x 2, y2) = (6.6)

Example # 2:

Use the remote to find the distance between (17,12) and form (9.6)

Let (x 1, y1) = (17,12)

Let (x 2, y2) = (9.6)

Example # 3:

Use the remote to find the distance between (17,12) and form (9.6)

Let (x 1, y1) = (7.8)

Let (x 2, y2) = (4.5)

Factoring the quadratic formula is the goal of this lesson. It is closely related to the resolution of equations using the quadratic formula

Step # 1:

Solve the quadratic equation for x 1 and x 2

Step # 2

Use the answers found in step # 1, the form of factoring is a (x – x 1) (x – x 2)

Example # 1:

Factor 4 x 2 + 9 x + 2 = 0, using the quadratic formula.

a = 4, b = 9, and c = 2

x = (BC ± v (b2 – 4ac)) / 2

x = (-9 ± v (92-4 × 4 × 2)) / 2 × 4

x = (-9 ± v (81-4 × 4 × 2)) / 8

x = (-9 ± v (81-4 × 8)) / 8

x = (-9 ± v(81-32)) / 8

x = (-9 ± v (49)) / 8

x = (-9 ± 7) / 8

x 1 = (-9 + 7) / 8

x 1 = (-2) / 8

x 1 =-1/4

x 2 = (-9-7) / 8

x 2 =-(16) / 8

x 2 = – 2

The factorization is a (x – x 1) (x – x 2)

The factorization is 4 (x – 1/4) (x – 2)

The factorization is 4 (x + 1/4)(x_+_2)

Now, to use distributive property to simplify the expression by getting rid of fractions

4 (x + 1/4)(x_+_2) = (4 × x + 4 × 1/4) (x + 2) = (4 x + 1) (x + 2)

Example # 2:

Factor x 2 + 2 x – 15 = 0, using the quadratic formula

a = 1, b = 2, c =-15

x = (BC ± v (b2 – 4ac)) / 2

x = (-2 ± v (22-4 × 1-15)) / 2 × 1

x = (-2 ± v (4-4 × 1-15)) / 2

x = (-2 ± v (× 4-4-15)) / 2

x = (-2 ± v (4 + 60)) / 2

x = (2 ± v (64)) / 2

x = (-2 ± 8) / 2

x 1 = (-2 + 8) / 2

x 1 = (6) / 2

x 1 = 3.

x 2 = (-2-8) / 2

x 2 = (-10) / 2

x 2 = – 5

The factorization is a (x – x 1) (x – x 2)

The factorization is 1 (x – 3) (x – 5)

The factorization is 1 (x – 3) (x + 5)

Now, use the distributive property to simplify the expression

1 (x – 3) (x + 5) = (1 × x + 1 /-× 3) (x + 2) = (x – 3) (x + 5)

It is important to understand how to use the quadratic formula front of fatoring using the quadratic formula.

## Distance calculator formula

The formula following distance calculator calculates the distance between two points on the coordinate system for you.

Guidelines for using the calculator

When you enter a number, do not use a bar slash: “/” or “”.

Point # 1: Enter item # 1 in the boxes to say x 1, y1.

Item # 2: Enter item # 2 in the boxes to say x 2, y2.

Do not enter units such as cm, inch, m, ft, etc….

Enter all the problems without unity. Then, after the calculator give you the answer, put the next response.

Example:

(x 1, y1) = (17 cm, 12 cm)

(x 1, y1) = (9 cm, 6 cm)

Enter 17 in the area who say x 1

Enter 12 in the area who say y1

Enter 9 in the box that say x 2

Enter 6 in the box who say y2

After you hit calculate, you should get 10 as a response

The answer is then 10 cm

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## Proof of the quadratic formula

The following is a proof of the quadratic formula. It will show you how the quadratic formula, which is widely used has been developed.

The evidence is performed using the standard form of a quadratic equation and the standard form of problems by completing the square

ax2 + bx + c = 0

Divide both sides of the equation by a kind you can fill the place

Subtract c/a leave both sides

Complete the square:

The coefficient of the second term is b / a

This coefficient divided by 2 and square the result to get (a b/2) 2

Add 2 (b/a) 2 on both sides:

Since the left side of the right side of the equation above is a perfect square, you can factor the left-hand side by using the coefficient of the first term (x) and the base of the last term(b/2a)

Add these two and raise at the second.

Then, square on the right side to get (b2) /(4a2)

Get the same denominator on the right side:

Now, take the square root of each side:

Simplify the left:

Rewrite the right side:

Subtract (b) / 2 (a) of both sides:

Adding the numerator, keeping the same denominator, we get the quadratic formula:

The + – between the b and the sign of the square root mean more or negative. In other words, most of the time, you will get two answers using the quadratic formula.

## Solve using the quadratic formula

This lesson shows how to solve using the quadratic formula. To use the quadratic formula, you must identify a, b and c in the standard of a quadratic equation

Quadratic formula is x = (BC ± v (b2 – 4ac)) / 2

The standard shape is ax2 + bx + c = 0

(1) for 6 x x 2 + 8 + 7 = 0, we get a = 6, b = 8, c = 7

(2) to x 2 + x 8 – 7 = 0, we get a = 1, b = 8, c =-7

(2) for x – 2-8 x + 7 = 0, we get a =-1, b =-8, c = 7

Example # 1:

Solve using the quadratic formula x x 2 + 8 + 7 = 0.

a = 1, b = 8, c = 7

x = (BC ± v (b2 – 4ac)) / 2

x = (-8 ± v (82-4 × 1 × 7)) / 2 × 1

x = (-8 ± v (64-4 × 1 × 7)) / 2

x = (-8 ± v (64-4 × 7)) / 2

x = (v(64-28) ±-8) / 2

x = (8 ± v (36)) / 2

x = (-8 ± 6) / 2

x 1 = (-8 + 6) / 2

x 1 = (-2) / 2

x 1 = – 1

x 2 = (-8-6) / 2

x 2 = (-14) / 2

x 2 = – 7

Example # 2:

Solve using the quadratic formula 4 x 2-11 x – 3 = 0

a = 4, b =-11 and c = – 3

x = (BC ± v (b2 – 4ac)) / 2

x = (-11 ± v ((-11) 2-4 × 4 ×-3)) / 2 × 4

x = (11 ± v (121-4 × 4-3)) / 8

x = (11 ± v (121-4 × – 12)) / 8

x = (11 ± v (121 + 48)) / 8

x = (± 11 v (169)) / 8

x = (11 ± 13) / 8

x 1 = (11 + 13) / 8

x 1 = (24) / 8

x 1 = 3.

x 2 = (11-13) / 8

x 2 = (-2) / 8

x 2 =-1/4

Example # 3:

Solve using the quadratic formula x 2 + x-2 = 0

a = 1, b = 1 and c =-2

x = (BC ± v (b2 – 4ac)) / 2

x = (-1 ± v (2-4 (1) × 1 ×-2)) / 2 × 1

x = (-1 ± v (1-4 × 1-2)) / 2

x = (-1 ± v (× 1-4-2)) / 2

x = (-1 ± v (1 + 8)) / 2

x = (1 ± v (9)) / 2

x = (-1 ± 3) / 2

x 1 = (-1 + 3) / 2

x 1 = (2) / 2

x 1 = 1.

x 2 = (-1-3) / 2

x 2 = (-4) / 2

x 2 = – 2