Solve: x cubed - 4x square - 21x = 0?

Question

I need the work to be shown on how answer was achieved.

Answer 1

Here ya go buddy. As x tends to infinity, only the highest power terms will dominate. All you have to do is compare the highest power terms in under each root. Effectively, the highest power of x in the first term is x^1 (since sqrt(x^2)=x) and the highest power in the second term is sqrt(x).

Now, as x->inf, you are effectively looking at the function x-sqrt(x). Since there are no denominator terms, this function goes straight to infinity.

Also, you could multiply by the conjugate to get
(x^2+7x+50)/((sqrt(x^2+8x+50)+...
but you'd still have to go through the same kind of rationale, where you would be dealing with:
lim x->inf x^2/x = x
which still goes to infinity. Only use the conjugate if you can't make this kind of connection or if you are evaluating your limit at a finite value.

Answer 2

First take x as a common factor:
x ( x^2 - 4x - 21) = 0
This means that either: x = 0 or x^2 - 4x - 21 = 0
The first equation is already done for you. The second needs the use of the quadratic formula. It gives you x = -3 or x = 7 (or if you don't know the quadratic forumal, you can factor the expression into: (x + 3)(x - 7) ).
So the solutions you are looking for are: x = 0, x = -3, or x = 7

Answer 3

x3 - 4x2 - 21x = 0. ("x3 - 4x2 - 21x" is a third degree polynomail)
x(x2 - 4x - 21) = 0. ("x2 - 4x - 21" is an Quadratic eqn.)
x(x2 - 7x + 3x - 21) = 0.
x(x(x-7) + 3(x-7)) = 0.
x(x+3)(x-7) = 0.

Now, since the original expression was of "degree 3", 'x' can have 3 values. Therefore, from above:

x=0 OR x=-3 OR x=7 - solution.
Solution Set (S.S) = {0, -3, 7}.

Answer 4

ax^2 +bx +c=0 purely use (-b+-SQRT(b-4ac))/2ac its common on a calculator that way in the different case do what speeded up at the same time promises c and extra at the same time promises b so for x^2+9x-10=0 -a million & 10 upload them at the same time promises 9 multiply you get -10 so the answer is (x-a million)(x+10)=0 now you may do x-a million=0 x+10=0 x=a million or x=-10 the different you ought to take the 4x^2 under consideration, that one you may do your self for samantha lower than, you want to divide each and everything by employing 9 so x^2 +9x = 10 must be (x^2)/9 +x = a million.11 so as that receives you nowhere

Answer 5

x^3 - 4.(x)^2-21x =0
x^2-4x-21=0
pattern = 21 result of multipliers 7 or -3
x^2-4x=21 x=7 and -3

Answer 6

x^3 - 4x^2 - 21x =
x(x^2 - 4x - 21) =
x(x - 7)(x + 3)

x = 0, 7, or -3

Answer 7

It's a quadratic, as you probably know.
First, pull out x because it's common in all the terms.
So you get x(x^2 -4x-21)=0
Then do the quadratic to get x times some other stuff. Then solve for x like normal.

Answer 8

x>3 - 4x>2 - 21x = 0
x(x>2 -4x -21) = 0
x(x +3)(x -7) = 0
x=0 or -3 or 7

Answer 9

first, factor out x to get
x(x^2-4x-21)=0
from there, factor to x(x-7)(x+3)=0
from there the answer should be pretty obvious

Answer 10

ill help you but will not solve the problem for you.

take x common at first.
then you have three terms in the brackets.
solve the quadratic equation.