Basic Algebra Solve for x

Question

Can someone provide the full steps solving for x:

2x^3 + 3x^2 -7 = 0

I'm looking to find the domain of the function and have computed from my calculator x= 1.149376...

So I know domain is (-infinity, 1.149376...) U (1.149376..., infinity)...just want to see full way to solve for x

Answer 1

I just want to specify that the number you got (1.149) is not the domain, it is a solution to the equation. Domain tells you what x values you can have and for all polynomials is (negative infinity, infinity). The solution, which is what you’ve found, tells you which value is an x intercept and that is x = 1.149. It’s picky but just be careful with words. To write (negative infinity, 1.149376) U (1.149376, infinity) as a domain would be incorrect because the function exists at x= 1.149 but your domain expression says we must jump over it. You would simply write x = 1.149 to give a solution.

This specific problem is actually impossible to solve by hand but I will work through our options to show you why. These may be steps you haven’t learned but helpful nonetheless.

First we are going to figure out how many real and imaginary zeroes there are. Remember, imaginary zeros come in conjugate pairs.

First, take your function and count how many times the terms change signs: this tells you the maximum possible real, positive zeros

2x^3 + 3x^2 -7 = 0

+ + -

There is one sign change, so there is max 1 possible real + zero

next take your function place (-x) where x is, and count how many times the terms change signs: this tells you the maximum possible real, negative zeros

2x^3 + 3x^2 -7 = 0

2(-x)^3 + 3(-x)^2 -7 = 0 simplify

-2x^3 + 3x^2 - 7 = 0

— +. -

2 sign changes, so there are a maximum of 2 real, negative zeroes

Lastly, we make a chart. We will have real +, real -, and imaginary as the headers, with all our combinations. We start with our maximum real zeroes and subtract 2 until we reach/pass 0, because imaginary zeroes come in pairs. Then, the imaginary zeroes make up the difference. If our leading term is 3rd degree, each row must add to 3.

Real +. Real - Imaginary

1. 2. 0. = 3

1. 0. 2. =. 3

As you can see, there was no way to subtract 2 from the positive real zero without passing 0. That zero is locked in; there is 1 positive real zero for certain, as we know will be 1.149, and either 2 negative zeroes or 2 imaginary zeroes.

The next step is to determine all the possible rational zeroes. Meaning, if this expression is factorable by hand, one of these numbers is a solution. The possible rational zeroes are:

factors of the constant / factors of the leading coefficient

These can be positive or negative!

factors of 7: +-1, +-7

factors of 2: +-1, +-2

Possible rational zeroes:

1/1 = 1

-1/1 = -1

1/2 = 1/2

-1/2 = -1/2

7/1 = 7

-7/1 = -7

7/2 = 7/2

-7/2 = -7/2

If your teacher intended this problem to be done by hand, one of these numbers would HAVE TO BE a solution.

Lastly, we test these numbers in one of two ways:

1) plug them into the equation for x; if the answer does come out to 0, then that number is a solution. We would then use synthetic division to remove that zero so we can look for other zeroes.

2) do synthetic division with these numbers as the outside number; if the remainder is 0, then that number is a solution, and what is left can be used to find remaining zeroes.

At this point, after exhausting through that list of numbers, you would find none of them to work. This would be the point when you realize your teacher intended you to solve this by graphing.

If one of those numbers did work, we would have a quadratic equation after doing synthetic division. We could easily find the remaining 2 zeroes by the quadratic formula. Then you would have an equation in fully factored form.

Looking at the graph we can see that x = 1.149 was the only place the graph touched the x axis. On our zeroes chart, this would mean the other 2 zeroes are imaginary and conjugates of each other. I did them out and they are -1.325+1.135i and -1.325 - 1.135i (rounded off) found using the quadratic formula after doing synthetic division with x = 1.149.

The main point is that it is only solvable by hand if one of the numbers in that list works. Otherwise, your teacher does not expect you to find it by hand because it’s simply not possible. Your only solution is x= 1.149, and the other 2 are not real. Hope this provides some insight!

Answer 2

There is some not too hard algebra that allows cubic equations to be

re-written as a multiplication of three linear polynomials.

if f(x)=ax³+bx²+cx+d

then also f(x)=a[(x-x₁)(x-x₂)(x-x₃)],,,,where x₁,x₂,and x₃ make f(x)=0

the x values can be calculated "by hand"

In this problem have 2[x³+(3/2)x²-7/2)=f(x)

let x=z-3/6,,,,,,,-3/6 = -(3/2)/3 which is negative squared x coefficient divided by 3

x=z-.5

now have 2[z³-(3/4)z-13/4]=f(z)

which can be re-written 2[(z-z₁)(z-z₂)(z-z₃)=f(z)

A mathematician Viete in the 1500's came up these formulas

the z coefficient is labeled 3Q, and the constant -2R ,,,,,after equation is reduced by x=z-.5

here Q=-1/4 and R=13/8

z₁=(R+√(R²+Q³))^(1/3)+(R-√(R²+Q³))^(1/3)

z₁=1.65

x₁=z₁-.5=1.15

z₂,z₃=(1/2)*[-z₁+/-(√-3)*√(z₁²+4Q)]

z_{2 and} z₃ in this case will be complex numbers √-1= i

z₂=(1/2)*(-1.65+i√(3*(1.65²-1))=-.825+1.14i

z₃=(1/2)*(-1.65-i√(3*(1.65²-1))=-.825-1.14i

and obviously

x₂=z₂-.5=-1.325+1.14i

x₃=z₃-.5=-1.325-1.14i