How do you find the exact solutions to the system y=x+3 and y=2x^2?

1 Answer
Oct 14, 2016

(3/2,9/2) and (-1,2)

Explanation:

You have to equal the two Ys, meaning their values as well or you can find the value of the first x and then plug it in the second equation. There are many ways to solve this.

y=x+3 and y=2x^2

y=y=>x+3=2x^2=>2x^2-x-3=0

You can use any tools you know to solve this quadratic equation but as for me, I will use Delta

Delta=b^2-4ac, with a=2, b=-1 and c=-3

Delta=(-1)^2-4(2)(-3)=25=> sqrt Delta=+-5

x_1=(-b+sqrt Delta)/(2a) and x_2=(-b-sqrt Delta)/(2a)

x_1=(1+5)/(4)=6/4=3/2 and x_2=(1-5)/(4)=-1

x_1=3/2 and x_2=-1

To find y, all you have to do is to plug the x values in either of the two equations. I will plug in both just to show you that it does not matter which one you chose.

With the first equation y=x+3

For x=3/2=>y=3/2+3=(3+6)/2=9/2

For x=-1=>y=-1+3=2

With the second equation y=2x^2

For x=3/2=>y=2(3/2)^2=1 color (red) cancel 2(9/(2 color (red) cancel4))=9/2

For x=-1=>y=2(-1)^2=2

Therefore, your solution is (3/2,9/2) and (-1,2)

Hope this helps :)