What is the solution set for x^3-3x^2+10x=30 (I know that x=3 but I need the solution)?

Answer 1

rewriting the equation

x^3 - 3x^2 + 10x - 30 = 0
then group the first two terms and the last two terms
(x^3 - 3x^2 ) + ( 10x - 30 ) = 0
now factor each group
x^2 ( x - 3 ) + 10 ( x - 3 ) =0
now factor the x - 3
(x - 3 ) ( x^2 + 10 ) = 0
set each factor equal to zero and solve

x - 3 = 0 and x^2 + 10 = 0
x = 3 and x^2 = - 10
sqrt x^2 = (+ or -) sqrt -10
x = (+ or -) i sqrt 10

those are the three roots, one real and two imaginary.

Answer 2

First: set the equation to "0" > subtract "30" from both sides....

x^3 - 3x^2 + 10x - 30 = 30 - 30
x^3 - 3x^2 + 10x - 30 = 0

Sec: when you have 4 terms, group "like" terms and factor....

(x^3 - 3x^2) + (10x - 30) = 0
x^2(x - 3) + 10(x - 3) = 0

(x - 3)(x^2 + 10) = 0

Third: you have two x-variables to solve for > set both parenthesis to "0"....

1. x - 3 = 0
*Add "3" to both sides....

x - 3 + 3 = 0 + 3
x = 3

2. x^2 + 10 = 0
*Subtract "10" from both sides....

x^2 + 10 - 10 = 0 - 10
x^2 = - 10

*Eliminate the exponent > find the square root of both sides....

V`(x^2) = +/- V`(- 10)

*You can't find the square root of a negative number, the negative sign becomes an imaginary number, represented as the variable "i" > factor out the "i"

x = +/- i V`(10) but, once you replace it with "x" in the original equation, the solution won't equal "0"

Solution: 3

Answer 3

x³ - 3x² + 10x = 30

x³- 3x² + 10x = 30

x³ - 3x² + 10x - 30 = 30 - 30

x³ - 3x² + 10x - 30 = 0

x²(x - 3) + 10(x - 3)= 0

(x² + 10)(x - 3) = 0

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Roots

x² + 10 = 0

x² + 10 - 10 = 0 - 10

x² = - 10

√x² = √- 10

x =√- 10. .<= . .Not used

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Roots

x - 3 = 0

x - 3 + 3 = 0 + 3

x = 3

The answer x = 3

Insert the x value into the equation

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Check

x³ - 3x² + 10x = 30

(3)³ - 3(3)² + 10(3) = 30

27 - 3(9) + 30 = 30

27 - 27 + 30 = 30

0 + 30 = 30

30 = 30

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Answer 4

Subtract 30 from both sides to get:

x^3 - 3x^2 + 10x - 30 = 0

This is a cubic equation.

Now factorise this. Note that (x-3) goes into the first two terms, and also into the last two terms.

So we can write the equation as:

(x - 3)(x² + 10) = 0

So (x-3) = 0 or (x² + 10) = 0

These are easy to solve.

Answer 5

take f(x)=x^3-3x^2+10x-30
x=3,f(3)=27-27=30-30=0
x-3, is a fraction of f(x)
(x-3)(x^2+10)=0
x=3,x^2+10=0the roots are complex

Answer 6

(x-3) * (x^2+10) = 0

x= -3, 10^1/2 * i