Question

Question #cfaee

Answer 1

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`

Answer 2

`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.