Solve the quadratic equation $(2-y)^4=3(2-y)^2+1$

Question

Solve $$(2-y)^4=3(2-y)^2+1$$

The answer is supposed to be $y=4\pm \sqrt{6+\frac{13}2}$. I have tried to work this problem out but I cannot get the answer that is in the book.

Answer 1

None of the 'book answers' you have provided represents a solution, instead the value listed at the end of these steps work correctly. I have listed my steps below. If you spot an error, please let me know:

1. We have: $$(2-y)^4=3(2-y)^2+1$$

2. let $$u=(2-y)^2$$

3. Substituting in original equation, you get: $$u^2-3u-1=0 $$

Using quadratic formula to solve the equation for u, $$u=\frac{3\pm \sqrt{9+4}}{2}$$

That is: $$u=\frac{3\pm \sqrt{13}}{2}$$

So far we found $u$, but we want to solve for $y$.

4. Equation in step 1 can be written as: $$u=(2-y)^2=4-4y+y^2$$

5. Which simplifies to:

$$y^2-4y+(4-u)=0$$

6. Note that we have already found the values of $u$, and we are solving for $y$ so we can apply the quadratic formula again, with y as a variable and u as a constant this time to get:

$$y=\frac{4\pm \sqrt{16-4(4-u)}}{2}$$

7. The above can be reduced to:

$$y=2\pm \sqrt{u}$$

8. We know the value of $u$ from step 3 above, so let's plug it in the above equation to get:

$$y=2\pm \sqrt{\frac{3\pm \sqrt{13}}{2}}$$

9. this results in the following values for $y$: $$y=3.817354021 , 0.182645979$$

This is the curve for: $$f(y)=(2-y)^4-3(2-y)^2-1$$