- #1

steem84

- 13

- 0

I am a little bit confused about solving the following equation:

(x-1)

How to do this??

(x-1)

^{2}(a+x)=1How to do this??

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- Thread starter steem84
- Start date

In summary, The original equation is (x-1)2(a+x)=1 and the solution can be found in figure 2. To solve the equation analytically, you can start by squaring both sides and simplifying to (\rho^{*})^6+((x^{*})^2-4a^2)(\rho^{*})^4+4a^2(a^2-(x^{*})^2)(\rho^{*})^2=0. Using the quadratic equation, you can solve for (\rho^{*})^2 and then take the square root to obtain the final solution.

- #1

steem84

- 13

- 0

I am a little bit confused about solving the following equation:

(x-1)^{2}(a+x)=1

How to do this??

(x-1)

How to do this??

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- #2

NoMoreExams

- 623

- 0

Are you sure there's a simple way to solve that, given that you don't know a? It's a cubic...

- #3

- #4

gabbagabbahey

Homework Helper

Gold Member

- 5,000

- 7

steem84 said:well, actually this is the original equation (see figure1)

The solution is in figure 2..

So let me reformulate my question: Can this be proven analytically?

This is very different from the problem in your first post!

Anyways, start by squaring both sides of the equation in the first link. Then multiply both sides by [tex]4((\rho^{*})^2+(x^{*})^2)[/tex] and simplify to obtain:

[tex](\rho^{*})^6+((x^{*})^2-4a^2)(\rho^{*})^4+4a^2(a^2-(x^{*})^2)(\rho^{*})^2=0[/tex]

That should tell you that either [tex](\rho^{*})^2=0[/tex]

[tex](\rho^{*})^4+((x^{*})^2-4a^2)(\rho^{*})^2+4a^2(a^2-(x^{*})^2)=0[/tex]

You can use the quadratic equation to solve the above expression for [tex](\rho^{*})^2[/tex] and then take the square root to obtain the final solution.

- #5

steem84

- 13

- 0

Yes, ok thank you. It did not cross my mind to substitute a variable for a variable^2

The expression (x-1)^2 * (a+x) can be simplified by using the distributive property. First, multiply (x-1)^2 to get x^2 - 2x + 1. Then, distribute this to (a+x) to get a(x^2) + (ax) - 2x^2 + x + 1. Finally, combine like terms to get a(x^2) - x^2 + (ax + x) + 1, which can be further simplified to (a-1)x^2 + (a+1)x + 1.

To solve for x, we need to isolate it on one side of the equation. First, distribute (x-1)^2 to get (a-1)x^2 + (a+1)x + 1 = 1. Then, subtract 1 from both sides to get (a-1)x^2 + (a+1)x = 0. We can then factor out x to get x((a-1)x + (a+1)) = 0. From here, we can use the zero product property and set each factor equal to 0 to solve for x. This will give us two possible solutions for x.

Yes, this equation can have complex solutions. When solving for x, we may end up with a quadratic equation that has complex roots. This means that the solutions for x will involve complex numbers, such as imaginary numbers.

Yes, you can check your solution by plugging it back into the original equation and solving for both sides. If the solution satisfies the equation, then it is correct. However, if it does not, then you may have made a mistake in your calculations.

Yes, this equation can be solved using the quadratic formula. After simplifying the expression and setting it equal to 0, we can compare it to the general quadratic equation, ax^2 + bx + c = 0. From there, we can use the quadratic formula to solve for x. However, it may be easier to factor the expression if possible before using the quadratic formula.

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