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

`3x^2-5x=-8`

`2x^2=6x-5`

`-x^2-10x=34`

Explanation:

The equations are:

`3x^2-5x=-8`

`2x^2=6x-5`

`12x=9x^2+4`

`-x^2-10x=34`

(a)
`3x^2-5x=-8`

Add 8 to both sides

`3x^2 - 5x + 8 =0`

To do this, we simply calculate the discriminant (d) using:

`d =b^2 - 4ac`

If
`d < 0`

Then it has complex roots

Where:
`a = 3; b = -5; c = 8`

`d = (-5)^2 - 4 * 3 * 8`

`d = 25 - 96`

`d = -71`

`-71 < 0` --- complex root

(b)
`2x^2=6x-5`

Equate to 0

`2x^2 - 6x + 5 = 0`

`d =b^2 - 4ac`

`d = (-6)^2 - 4 * 2 * 5`

`d = 36 - 40`

`d =-4`

`-4 < 0` --- complex root

(c)
`12x=9x^2+4`

Equate to 0

`9x^2 - 12x + 4 = 0`

`d =b^2 - 4ac`

`d = (-12)^2 - 4 * 9 * 4`

`d = 144 - 144`

`d = 0` ---- real roots

(d)
`-x^2-10x=34`

Equate to 0

`-x^2 - 10x - 34 = 0`

`d =b^2 - 4ac`

`d = (-10)^2 - 4 * (-1) * (-34)`

`d = 100 - 136`

`d = - 36`

`-36 < 0` ---- complex root