x^3 + 9x =6x^2 solve by factoring?

Question

solve the equation, noting any double roots. ineed it as soon as possible.

Answer 1

x^3+9x=6x^2
=>x^3-6x^2+9x=0
=>x(x^2-6x+9)=0
=>x(x-3)^2=0
Therefore either x=0 or 3

Answer 2

this isn't a lot factoring with the aid of the indisputable fact that's plug and examine. purely the right answer holds real equations for the x pronounced. once you discover the right answer, you do have got here upon the factoring to boot (as i will educate contained in the first answer): a million:3x2 + 7x = -2 -2, -a million/3 --> 3x2 + 7x + 2 = (x+2)*(x+a million/3)*3 2: x2 - 40 9 = 0 -7, 7 3: x2 + 5x + 4 = 0 -a million, -4 4: 6x2 - 2 = x -a million/2, 2/3 5: x2 - x = 40 2 7, -6 6: x2 - 21x - seventy 2 = 0 24, -3 7: 7x2 - 7x - 2x + 2 = 0 2/7, a million 8: 3x2 - 2x - 5 = 0.5/3, a million 9: 2a2 - 9a = 5 -a million/2, 5 10: 33x2 - x - 14 = 0 2/3, -7/11 11: 21x2 + 22x - 24 = 0 2/3, -12/7 12: 2x2 + 2x = 60 5, -6 13: x2 + 16x + 40 8 = 0 -4, -12 14: x3 - x2 - 12x = 0 -3, 4, 0 15: 4x2 + 8x + 4 = 0 a million, 2

Answer 3

x^3 + 9x =6x^2 set it equal to zero

x^3 - 6x^2 + 9x = 0, see what you can factor out.

x ( x^2 - 6x +9 ) = 0, x factors out. This take care of the x^3.

x (----)(----) = 0

x(x---)(x---) = 0, this takes care of the x^2

x(x---)(x---) = 0, what two numbers add to be -6, and multiplied make 9. It would be -3

x(x-3)(x-3)=0, set each equal to zero, and solve.

x=0,
x-3 = 0 + 3 , so x = 3
x-3 = 0 + 3 , so x = 3

3 is a double root

Answer 4

First, I suggest you set the whole equation equal to 0, and write in descending order..

x^3 - 6x^2 + 9x = 0

Now, factor out the greatest common factor (always do this first.)

x(x^2 -6x + 9) = 0

Looking at the second factor, you may notice that it fits the pattern for a perfect square trinomial (square root of the quadratic coefficient times the square root of the constant, times 2). Thus, you get

x(x - 3)^2 = 0.

Now, use the Zero Product rule that states "If you have a number of factors whose product is 0, then one of them must be 0." Set each factor equal to 0.

x = 0, (x - 3)^2 =0

Then, solve for x:

x = 0 is solved already;

(x - 3)^2 = 0
x - 3 = 0 (by taking square root of both sides)
x = 3

So, your answers are

x = 0

and

x = 3, with a multiplicity of 2 (that means it is a root twice.)

Answer 5

x^3 + 9x =6x^2

from here you take an x out of the entire problem...

x(x^2-6x+9)=0

Now you divide zero by x which is zero

solution #1 = 0

you are left with the equation x^2-6x+9=0

this factors to (x-3)(x-3)

set the seperate parts of the monomial = to 0

x=3
and
x=3

solution #1=0

#2=3

#3=3

Answer 6

Rewrite this as x(x^2 - 6x +9) = 0, and see at once that the roots are 0, 3, and 3.

Answer 7

rewrite x^3 -6x^2 +9x =0

put x in factor x(x^2-6x+9) =0

Notice that (x^2-6x+9) = (x-3)^2

so you have x (x-3)^2 =0

2roots a simple x=0 and a double one x = +3

Answer 8

we can rewrite the eqn as
x^3-6x^2+9x=0
taking common term x out we have
x(x^2-6x+9)=0
x(x-3)(x-3)=0 using formula a^2-2ab+b^2=(a-b)^2 for( x^2-6x+9)

so x=0,3

Answer 9

x^3 - 6x^2 + 9x=0
x (x^2 - 6x + 9) = 0
x (x^2 - 3x - 3x + 9) = 0
x [x(x - 3) - 3(x - 3)] = 0
x (x-3) (x-3) = 0
x = 0 or x = 3

Answer 10

x^3 - 6x^2 + 9x=0
x (x^2 - 6x + 9) = 0
x (x^2 - 3x - 3x + 9) = 0
x [x(x - 3) - 3(x - 3)] = 0
x (x-3) (x-3) = 0
x = 0 or x = 3