Show that... Answer confermation please:)

12345678

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Mar 30, 2013
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‘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.

 
Looks good though all I'd say is that the discriminant is always non-negative.

That is not sufficient. That leaves open the possibility that the discriminant might/could be zero, and then
there would not be two distinct solutions as was required as part of the original problem.
 
I'll bite: how can a^2 + 40 = zero?

Only if a = SQRT(-40) , right?
No one said that. Lookagain's point was just that saying "because the determinant is non-negative" is not sufficient. You must say "because the determinant is postive" which, of course, it is here.
 
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