Question

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

A 3x>2-5=-8
b 2x>2=6x-5
c 12x=9x>2 +4
d -x>2 - 10x =34

Answer 1

see details below

Explanation:

First, assume >2 was meant to be &super;. or square.

then the given equations are:

A 3x^2-5=-8

b 2x^2=6x-5

c 12x=9x^2 +4

d -x^2 - 10x =34

In all cases, rearrange equation to identify parameters a, b and c, then calculate determinant using D=b^2-4ac.

A 3x^2x-5=-8

3x^2-5x+8=0, a=3, b=-5, c=8

D=b^2-4ac = (-5)^2-4*3*8 = 25 - 96 = -71 < 0 => 2 complex solutions

B 2x^2=6x-5

2x^2-6x+5, a=2, b=-6, c=5

D=b^2-4ac = (-6)^2-4*2*5 = 36 - 40 = -4 < 0 => 2 complex solutions

C 12x=9x^2 +4

9x^2-12x+4, a = 9, b=-12, c=4

D=b^2-4ac = (12)^2-4*9*4 = 144 - 144 = 0 => 2 real coincident solutions

D -x^2 - 10x =34

-x^2-10x-34, a = -1, b=-10, c=-34

D=b^2-4ac = (-10)^2-4*(-1)*(-34) = 100 - 136 = -36 <0 => 2 complex solutions