What are the solutions of the system?

y=x^2+4.6x+4.7
y=3.3x-3.7

A. (–3.47, 3.48) and (2.17, –15.17)

B. (–3.47, –15.17) and (–2.17, –10.88)

C. (–3.47, –15.17) and (2.17, 3.48)

D. no solution

A, B, C, or D? Please explain :)

Question

What are the solutions of the system?

y=x^2+4.6x+4.7
y=3.3x-3.7

A. (–3.47, 3.48) and (2.17, –15.17)	

B. (–3.47, –15.17) and (–2.17, –10.88)	

C. (–3.47, –15.17) and (2.17, 3.48)	

D. no solution	

A, B, C, or D? Please explain :)

kira_yamato · Answer

Equate the RHS of both together.
x^2 + 4.6x + 4.7 = 3.3x - 3.7
x^2 + 1.3x - 1 = 0
Using quadratic formula
$$x = \frac{-1.3 ± \sqrt{(1.3)^2 - 4(1)(8.4)}}{2(1)} = \frac{-1.3 ± \sqrt{-31.91}}{2}$$
No soln

anonymous · Answer

Are you positive that there is no solution?

anonymous · Answer

Well then, thank you for all of your help.

kira_yamato · Answer

Because apparently there's no real solution... Because you cannot get a real no. by sqrt -ve no.