how many possible solutions are there in the system x^2+2x+2y=5 and 3x^2-4x=3+y?
y = 3 x^2 - 4 x - 3
so put that in the top one
x^2 + 2x + 2 [ 3 x^2 - 4 x - 3 ] = 5
x^2 + 2x + 6 x^2 - 8 x - 6 = 5
7 x^2 - 6 x - 11 = 0
b^2-4 a c = 36 - 4*77 very negative gives two COMPLEX, no real, roots
x² + 2 x + 2 y = 5
3 x² - 4 x = 3 + y
Subtract 3 to both sides
3 x² - 4 x - 3 = y
Replace y = 3 x² - 4 x - 3 in equation x² + 2 x + 2 y = 5
x² + 2 x + 2 ( 3 x² - 4 x - 3 ) = 5
x² + 2 x + 6 x² - 8 x - 6 = 5
7 x² - 6 x - 6 = 5
Subtract 5 to both sides
7 x² - 6 x - 11 = 0
Discriminant:
D = b² - 4 a c = ( - 6 )² - 4 ∙ 7 ∙ ( - 11 ) = 36 + 308 = 344
Discriminant is positive so there are two REAL roots..
In this case the solutions are:
x1 = ( 3 - √86 ) / 7 and x2 = ( 3 + √86 ) / 7