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

Answer:

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

$2x^2=6x-5$

$-x^2-10x=34$

Step-by-step 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