60.9k views
4 votes
PLEASE HELP ASAP

Which equations have no real solution but have two complex solutions?

3x^2-5x=-8 2x^2=6x-5 12x=9x^2+4 -x^2-10x=34

1 Answer

5 votes

Answer:

First option:
3x^2-5x=-8

Second option:
2x^2=6x-5

Fourth option:
-x^2-10x=34

Explanation:

Rewrite each equation in the form
ax^2+bx+c=0 and then use the Discriminant formula for each equation. This is:


D=b^2-4ac

1) For
3x^2-5x=-8:


3x^2-5x+8=0

Then:


D=(-5)^2-4(3)(8)=-71

Since
D<0 this equation has no real solutions, but has two complex solutions.

2) For
2x^2=6x-5:


2x^2-6x+5=0

Then:


D=(-6)^2-4(2)(5)=-4

Since
D<0 this equation has no real solutions, but has two complex solutions.

3) For
12x=9x^2+4:


9x^2-12x+4=0

Then:


D=(-12)^2-4(9)(4)=0

Since
D=0 this equation has one real solution.

4) For
-x^2-10x=34:


-x^2-10x-34=0

Then:


D=(-10)^2-4(-1)(-34)=-36

Since
D<0, this equation has no real solutions, but has two complex solutions.

User DefyGravity
by
5.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.