Which system of equations does not have a real solution?
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

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.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb