Question

. Having survived the meteor impact, thanks to some last minute evasive maneuvers, the Mathonauts now set their sights on their Interstellar Headquarters. The Interstellar Headquarters orbits the Earth based on the equation y2 + x2 = 40,000. Using the original trajectory of the ship and complete sentences, explain to the pilot how to find where the ship’s path will cross the Interstellar Headquarters’s path.

Answer 1

My original equation is y=2x+4

Answer 2

What I thought would be the first step in this process is that I first square root the equation y2 + x2 = 40,000. correct @mathstudent55

Answer 3

You can't take the square root of the second equation.
You need to solve the two equations simultaneously.

Answer 4

You can use the substitution method.
Since the equation y = 2x + 4 is already solved for y, substitute y in the other equation with y^2 + x^2 = 40,000.

Answer 5

Ok what that would mean is that I plug in the y value I had gotten which is 2x+4 and square that which would cause for me to foil

Answer 6

Yes.

Answer 7

\(y = 2x + 4\)

\(y^2 + x^2 = 40000\)

Substitute y in the second equation with \(2x + 4\).

\((2x + 4)^2 + x^2 = 40000\)

Answer 8

If I foiled it I got 4x^2+16x^2+16

Answer 9

Which if I simplified again i GOT 18X^2

Answer 10

Which the equation I got was 18x^2+16

Answer 11

@mathstudent55

Answer 12

This is how you square a binomial:

\((a + b)^2 = a^2 + 2ab + b^2\)

In your case, you get:

\((2x + 4)^2 + x^2 = 40000\)

\(4x^2 + 16x + 16 + x^2 = 40000\)

Answer 13

Using FOIL you got it almost right.
The 16x term is just 16x, not 16x^2.

Answer 14

Do you understand it so far?

Answer 15

Yes

Answer 16

Using FOIL, squaring your binomial is like this:

\((2x + 4)(2x + 4) = 4x^2 + 8x + 8x + 16 = 4x^2 + 16x + 16\)

Answer 17

Ok.
Now we take the equation, and we combine like terms.

Answer 18

\(4x^2 + 16x + 16 + x^2 = 40000\)

4x^2 and x^2 are like terms.
Also, we subtract 40000 from both sides.

\(4x^2 + x^2 + 16x + 16 - 40000 = 0\)

\(5x^2 + 16x - 39984 = 0\)

Answer 19

Now we need to solve that quadratic equation for x.

Answer 20

ok

Answer 21

We need to use the quadratic formula.
Are you familiar with it?

Answer 22

yes I am

Answer 23

-b^2+squareroot of b^2-4ac/2a
-

Answer 24

The solution of the quadratic equation \(ax^2 + bxc + c = 0\) is

\(x=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Answer 25

Be careful. The first term is simply b, not b^2.

Answer 26

yeah

Answer 27

\(x = \dfrac{-16 \pm \sqrt{16^2 - 4(5)(-39984)}}{2(5)}\)

Answer 28

I got -16+ squaroot of 79993.6
-

Answer 29

\(x = \dfrac{-16 \pm \sqrt{256 + 799680}}{10}\)

\(x = \dfrac{-16 \pm \sqrt{799936}}{10}\)

\(x = \dfrac{-16 \pm \sqrt{799936}}{10}\)

Answer 30

Now we need to simplify the root.

\(x = \dfrac{-16 \pm \sqrt{64 \times 12499}}{10}\)

Answer 31

\(x = \dfrac{-16 \pm 8\sqrt{12499}}{10}\)

Answer 32

\(x = \dfrac{-8 \pm 4\sqrt{12499}}{5}\)

\(x = \dfrac{-8 + 4\sqrt{12499}}{5}\) or \(x = \dfrac{-8 - 4\sqrt{12499}}{5}\)

Answer 33

Now we plug in each x-value we got in the first equation, and we find each corresponding y-value.

Answer 34

\(x = \dfrac{-8 + 4\sqrt{12499}}{5}\) or \(x = \dfrac{-8 - 4\sqrt{12499}}{5}\)

\(y = 2x + 4\)

\(y = 2\times \dfrac{-8 + 4\sqrt{12499}}{5} + 4\); \(y = 2\times \dfrac{-8 - 4\sqrt{12499}}{5} + 4\)

\(y = \dfrac{-16 + 8\sqrt{12499}}{5} + \dfrac{20}{5}\); \(y = \dfrac{-16 - 8\sqrt{12499}}{5} + \dfrac{20}{5}\)

\(y = \dfrac{4 + 8\sqrt{12499}}{5}\); \(y = \dfrac{4 - 8\sqrt{12499}}{5}\)

The solutions are:

\(x = \dfrac{-8 + 4\sqrt{12499}}{5}\), \(y = \dfrac{4 + 8\sqrt{12499}}{5}\)

\(x = \dfrac{-8 - 4\sqrt{12499}}{5}\), \(y = \dfrac{4 - 8\sqrt{12499}}{5}\)

Answer 35

Thank you