In the system shown below, what are the coordinates of the solution that lies in quadrant II?
x^2+4y^2=100
y= (1/32)x^2

Question

In the system shown below, what are the coordinates of the solution that lies in quadrant II?
x^2+4y^2=100
y= (1/32)x^2

mathstudent55 · Answer

Can you solve the second equation for x^2?

anonymous · Answer

umm no.. im not that bright

mathstudent55 · Answer

Don't say that. We'll just do it together.
Since the first equation has x^2 and y^2 in it, if we solve the second equation for x^2, then we can substitute it into the first equation.

anonymous · Answer

lol thanks

mathstudent55 · Answer

Notice the right side of the second equation is (1/32)x^2
Since x^2 is being multiplied by 1/32, if we multiply the right side by the reciprocal of 1/32 (which is 32), we'll get rid of the 1/32 leaving just x^2, what we want.

Remember, that in an equation you must do the same thing to both sides, so we start by multiplying both sides of the second equation by 32.

anonymous · Answer

so y=x^2 is left..

mathstudent55 · Answer

$y= \dfrac{1}{32} x^2 $

$\color{red}{32 \times} y= \color{red}{32 \times} \dfrac{1}{32} x^2 $

mathstudent55 · Answer

You must also multiply the left side by 32.

mathstudent55 · Answer

$32y = x^2$

or

$x^2 = 32y$

anonymous · Answer

i am totally not being rude... but i only have 16 minutes left to answer 23 more questions -_-

anonymous · Answer

36y^2 =100 is left.. ?

anonymous · Answer

i really appreciate your help though!! :) i have 12 minutes left.. lol

anonymous · Answer

i have 8 minutes left and it wont let me move on until i answer this one.. can you please hurry -_-

mathstudent55 · Answer

Now that we know what x^2 is equal to, we substitute it in the first equation:

$x^2+4y^2=100$

$32y + 4y^2 = 100$

$8y + y^2 = 25$

$y^2 + 8y - 25= 0$

 $y = \dfrac{-8 \pm \sqrt{8^2 - 4(1)(-25)}}{2(1)} $

$y = \dfrac{-8 \pm \sqrt{64 + 100}}{2} $

$y = \dfrac{-8 \pm \sqrt{164}}{2} $

$y = \dfrac{-8 \pm 2\sqrt{41}}{2} $

$y = -4 \pm \sqrt{41} $

Since we only want quadrant one values, we keep

$y = -4 + \sqrt{41} $

mathstudent55 · Answer

Now we plug this value into the second equation and find a positive value of x.

$y = \dfrac{1}{32}x^2$

$-4 + \sqrt{41} = \dfrac{1}{32} x^2 $

$-128 + 32\sqrt{41} = x^2$

$x = \pm \sqrt{-128 + 32\sqrt{41}} $

Discard the negative solution, and simplify the radical.