There is a point $A$ with positive coordinates such that the sum of the coordinates of $A$ is $14$. If the $x$-coordinate of $A$ is $a$, then point $P$ is at $(3a, a^2+13a-11)$. If the slope of a line passing through $A$ and $P$ is $7$, find $a$.

The x-coordinate of A is:

a = 6.1

Step-by-step explanation:

For point A:

x-coordinate = a

x-coordinate + y-coordinate = 14

y-coordinate = 14 - x-coordinate

y-coordinate = 14 - a

So point A can be written as (a, 14-a)

Let (a,14) be (x₁,y₁)

Point P can be written as (3a, a²+13a-11)

Let P be (x₂,y₂)

Slope of a line passing through 2 point is given as:

where m=7

Substitute all values

As A has only positive coordinates

a = 6.1

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.