If you mean b^2 -4ac then it is the discriminant of a quadratic equation. If the discriminant equals 0 then the equation has 2 equal roots. If the discriminant is greater than 0 then the equation has 2 different roots. If the discriminant is less than 0 then it has no real roots.
x2 + 3x - 18 = 0 (x - 3)(x + 6) = 0 x = 3 or -6
x2-14x+45=0(x-9)(x-5)=0x-9=0 & x-5=0x=9 & x=5
A linear equation.
Using the discriminant the possible values of k are -9 or 9
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.
-4,3 are the roots of this equation, so for the values for which the sum of roots is 1 & product is -12
If you mean b^2 -4ac then it is the discriminant of a quadratic equation. If the discriminant equals 0 then the equation has 2 equal roots. If the discriminant is greater than 0 then the equation has 2 different roots. If the discriminant is less than 0 then it has no real roots.
It has roots x = 2.618 and x = 0.38197
The product of the roots of the equation 2x2 -x -2 = 2 is 2x2 -x -2 = 2.
The roots are: x = -5 and x = -9
The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842
This quadratic equation has no real roots because its discriminant is less than zero.
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.
x = 2 and x = 4
There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2
It is a quadratic equation in X, with two real roots.