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) = 4
Therefore, the roots are not imaginary.
B) x2 - 2 = 0
= 0 - 4(1)(-2) = 8
Therefore, the roots are not imaginary.
C) x2 + x + 1 = 0
= 1 - 4(1)(1) = -3
Therefore, the roots are imaginary.
D) x2 - x - 1 = 0
= 1 - 4(1)(-1) = 5
Therefore, the roots are not imaginary.
The equation x2 + x + 1 = 0 has imaginary roots.
In the context of algebra, the term real root refers to the solution to an equation which consists of a real number rather than an imaginary or complex number (a complex number being a combination of real and imaginary numbers). You may recall that any given equation will have the same number of roots (or solutions) as the highest exponent in the equation, so that if you are dealing with x squared, you have two roots. Often there would be one real root and one imaginary root. In general, the real roots are more useful, although there are some circumstances in which imaginary or complex roots are also relevant to what you are doingl.
It has roots x = 2.618 and x = 0.38197
When you need to find the roots of a quadratic equation and factorisation does not work (or you cannot find the factors). The quadratic equation ALWAYS works. And when appropriate, it will give the imaginary roots which, judging by this question, you may not yet be ready for.
This quadratic equation has no real roots because its discriminant is less than zero.
The roots are: x = -5 and x = -9
roots
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
No real roots. Imaginary roots as this function does not intersect the X axis.
5n2 + 2n + 6 = 0 is a quadratic equation in the variable n.The equation does not have any real roots. The roots are the complex conjugate pair -0.2 ± 1.077i where i is the imaginary square root of -1.
-4,3 are the roots of this equation, so for the values for which the sum of roots is 1 & product is -12
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.
In the context of algebra, the term real root refers to the solution to an equation which consists of a real number rather than an imaginary or complex number (a complex number being a combination of real and imaginary numbers). You may recall that any given equation will have the same number of roots (or solutions) as the highest exponent in the equation, so that if you are dealing with x squared, you have two roots. Often there would be one real root and one imaginary root. In general, the real roots are more useful, although there are some circumstances in which imaginary or complex roots are also relevant to what you are doingl.
They are called the solutions or roots of the equations.
the expression "b2-4ac" with respect to quadratic equations is called the discriminant. the discriminant of the equation tells whether or not the roots will be real numbers or not. If the discriminant is negative, then the roots are imaginary.
The product of the roots of the equation 2x2 -x -2 = 2 is 2x2 -x -2 = 2.
It has roots x = 2.618 and x = 0.38197
When you need to find the roots of a quadratic equation and factorisation does not work (or you cannot find the factors). The quadratic equation ALWAYS works. And when appropriate, it will give the imaginary roots which, judging by this question, you may not yet be ready for.