Which system of equations does not have a real solution
Asked by maham237 @ in Mathematics viewed by 301 People
A. y = x 2 + 3x - 5 and x + y = -10
B. y = x 2 + 3x - 5 and 4x + 5y = 20
C. y = x 2+ 3x- 5 and x + y = -9
Answered by maham237 @
Answer:
It's A.
Step-by-step explanation:
Let's look at option A:
From the second equation y = -10 - x. Substituting in the first equation:
-10 - x = x^2 + 3x - 5
x^2 + 4x + 5 = 0
Checking the discriminant b^2 - 4ac we get 16 - 4*1*5 = -4 so there are no real roots. (A negative discriminant means no real roots).
So A has no real solution.
B.
x^2 + 3x - 5 = (20 - 4x)/5 = 4 - 0.8x
x^2 +3.8x - 9 = 0
b^2 - 4ac = (3.8)^2 - 4*1*-9 = 50.44 (positive) so there are real roots.
C.
x^2 + 3x - 5 = -9 - x
x^2 + 4x + 4 = 0
b^2 - 4ac = 4^2 - 4*1*4 = 0 so there are real roots.
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