There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
It has two complex roots.
To find which has imaginary roots, use the discriminant of the quadratic formula (b2 - 4ac) and see if it's less than 0. (The quadratic formula corresponds to general form of a quadratic equation, y = ax2 + bx + c)A) x2 - 1 = 0= 0 - 4(1)(-1) = 4Therefore, the roots are not imaginary.B) x2 - 2 = 0= 0 - 4(1)(-2) = 8Therefore, the roots are not imaginary.C) x2 + x + 1 = 0= 1 - 4(1)(1) = -3Therefore, the roots are imaginary.D) x2 - x - 1 = 0= 1 - 4(1)(-1) = 5Therefore, the roots are not imaginary.The equation x2 + x + 1 = 0 has imaginary roots.
x2 + 11x + 30 = 0 (x + 5)(x + 6) = 0 so the roots are -5 and -6
It has roots x = 2.618 and x = 0.38197
No real roots
The roots are: x = -5 and x = -9
Doesn't have integer roots. Quadratic formula gives roots as 3.71 and -5.38.
To the nearest hundredth, roots are -1.71 and 3.21
-4,3 are the roots of this equation, so for the values for which the sum of roots is 1 & product is -12
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
x2-4x+4 = 0 (x-2)(x-2) = 0 x = 2 or x = 2 It has two equal roots.