Question

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

Answer 1

Answer:

First option:  $3x^2-5x=-8$

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

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

Step-by-step 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.

Answer 2

Answer:

First option:  

Second option:

Fourth option:

Step-by-step explanation: