‘Show that the simultaneous equations:
2x² + xy = 10
x + y = a
Always have two distinct solutions, for all possible values of the constant a’
I was just wondering if my working actually shows this?
I started by rearranging x + y = a, in order to isolate y: y = a – x
I then subbed y = a – x into 2x² + xy = 10
= 2x² + x (a – x) = 10
=2x² + ax - x² = 10
=x² + ax = 10
= x² + ax – 10 = 0
Using the discriminant b² - 4ac, a = 1, b ≥1 c = -10
Discriminant = 1² - (4*1*-10) = 1 – (-40) = 41
As b≥1 and c is negative, the discriminant will always be greater than 0, meaning the root will always have 2 distinct solutions.
2x² + xy = 10
x + y = a
Always have two distinct solutions, for all possible values of the constant a’
I was just wondering if my working actually shows this?
I started by rearranging x + y = a, in order to isolate y: y = a – x
I then subbed y = a – x into 2x² + xy = 10
= 2x² + x (a – x) = 10
=2x² + ax - x² = 10
=x² + ax = 10
= x² + ax – 10 = 0
Using the discriminant b² - 4ac, a = 1, b ≥1 c = -10
Discriminant = 1² - (4*1*-10) = 1 – (-40) = 41
As b≥1 and c is negative, the discriminant will always be greater than 0, meaning the root will always have 2 distinct solutions.