Question

Give the point (or points) of intersection of x2 + y2 = 5 and y = 2/3x + 1 Round all decimal values to 2 places. Your answer must not be more than 0.01 from the true value to receive credit. Be careful with your rounding.

Answer 1

x² + y² = 5

y = 2x/3 + 1

In a simple way to find the intersections you can just throw the y value in the other equation, so

x² + y² = 5

x² + (2x/3 + 1)² = 5

x² + (2x/3)² + 2.2x/3.1 + 1² = 5

x² + 4x²/9 + 4x/3 + 1 = 5

13x²/9 + 4x/3 + 1 - 5 = 0

13x²/9 + 4x/3 - 4 = 0

Multiply by 9

13x² + 12x - 36 = 0

Δ = b² - 4.a.c

Δ = 12² - 4 . 13 . -36

Δ = 144 - 4. 13 . -36

Δ = 2016

There are 2 real roots.

x = (-b +- √Δ)/2a

x' = (-12 + √2016)/2.13

x'' = (-12 - √2016)/2.13

x' = 32,8998886412873 / 26

x'' = -56,8998886412873 / 26

x' = 1,2653803323572037

x'' = -2,188457255434127

Rounding

x' = 1,27

x'' = -2,19

Now we need to put this value in the y equation to know the y coordinate.

y = 2x/3 + 1

y' = 2.1,27/3 + 1

y' = 2,54/3 + 1

y' = 1,8466666...

y' = 1,85

y = 2x/3 + 1

y'' = 2.(-2,19)/3 + 1

y'' = -4,38/3 + 1

y'' = -0,46

So the points are: (1,27;1,85) and (-2,19;-0,46)