Answer:
Part 1) The quadratic equation has zero real solutions
Part 2) The solutions are
and
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
The discriminant is equal to

If D=0 -----> the quadratic equation has only one real solution
If D>0 -----> the quadratic equation has two real solutions
If D<0 -----> the quadratic equation has two complex solutions
<em>Find the value of D</em>
-----> the quadratic equation has two complex solutions
<em>Find out the solutions</em>
substitute the values of a,b and c in the formula
Remember that

Answer:
Similarly, the distance between two points P1 = (x1,y1,z1) and P2 = (x2,y2,z2) in xyz-space is given by the following generalization of the distance formula, d(P1,P2) = (x2 x1)2 + (y2 y1)2 + (z2 z1)2. This can be proved by repeated application of the Pythagorean Theorem.
Step-by-step explanation:
Use a calculator
Answer:
67.02
Step-by-step explanation:
Answer:
The zeros of f(x) are -3, 2 , 6
Step-by-step explanation:
f(x) is a polynomial of degree 3.
If the polynomial is not factorized we will either factorize to find the zero or use trial and error method.
Since the f(x) in the question is in the factorized form. we will have to equate each factor to zero.
f(x) = (x+3)(x-2)(x-6)
x + 3 = 0 => x = -3
x - 2 = 0 => x = 2
x - 6 = 0 => x = 6