Question #cfaee

2 Answers
Sep 11, 2017

Answer:

The possible values for #x# are 0, 8 or -3

Explanation:

What we have to do is, somehow, come up with an equation that resembles #ax^2+bx+c=0#, in our case we have:

#x^3-24x=5x^2#

first lets make it so one side equals zero:

#x^3-5x^2-24x=0#

Now, we can factor out one #x# out of each term, leaving us with:

#x*(x^2-5x-24)=0#

So, for any multiplication to be equal to zero, one of the terms is equal to zero. That leaves us with the two following conditions:

Either #x=0" "# or #" "x^2-5x-24=0#

Now we take our #a, b# and #c# (1, -5 and -24) from the second condition and plug them into the quadratic formula to find the other values of x:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(5+-sqrt((-5)^2-4(1)(-24)))/(2(1))#

#x=(5+-sqrt(25-(-96)))/2#

#x=(5+-sqrt(121))/2#

#x=(5+-11)/2#

#x=16/2=8" "# or #" "x=-6/2=-3#

Sep 11, 2017

Answer:

#x=8 or x=-3 or x =0#

Explanation:

Note : This is not a quadratic equation, a quadratic equation is of degree 2. Since this equation is of degree 3, it is correct to say that it is a cubic equation.

To factor, or factorise means to write an expression as the product of its prime factors.

Factoring #42: " " 42 = 2xx3xx7#

There are several ways to factorise:

  • divide out a common factor
  • divide out a common bracket
  • grouping terms
  • quadratic trinomial
  • difference of squares

We have #x^3 -24x= 5x^2#

Move all the terms to one side and make the other side #0#

#x^3 -5x^2 -24x =0" "larr# divide out #x# from each term

#x(x^2-5x-24) =0#

Find factors of #24# which differ by #5#

They are #8 and 3#

The signs will be different, the bigger is negative.
#-8 xx +3 = 24" " and -8+3 = -5#

#:.x(x-8)(x+3)=0#

There are #3# factors, so there will be #3# solutions.

Set each factor equal to zero:

#x =0#
#x-8=0 rArr x =8#
#x+3=0 rArr x =-3#

These are the 3 possible solutions.