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For example: in $(x-1)^2 + (y+2)^2 = 4$

I am in a region where BIMDAS is taught as the order of operations. My logic is to start by squaring because there is nothing that can be done inside of the brackets to begin with except simplifying (0 + 2), so I move onto dealing with indices by removing them via rooting both sides of the equation.

Solving for x at y = 0, if I start by taking the square root of both sides of the equations, I get principal root 2 and (secondary?) root -2 on the right hand side.

When I complete the calculation of x I end up with 1 and -3, but only answer (1) is an intersection of the x plane.

What is going on here?

If I begin by expanding brackets, I arrive at only one answer (1), which is correct.

duckegg
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    No. Taking square roots first does nothing for you. It is not the case that $\sqrt{a+b}$ is the same as $\sqrt{a} +\sqrt{b}.$ – Sean Roberson Jan 02 '22 at 23:48
  • Acronym BINDAS meaning ... – Jean Marie Jan 03 '22 at 00:00
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    You might be interested in posting mathematical notation to make your steps clear to your Readers. Keep your work simple and neat as possible. Since the $x$-intercept is where $y=0$, start by substituting that in the circles equation. – hardmath Jan 03 '22 at 00:32
  • Not an answer to your algebraic conundrum (@SeanRoberson has that) but think geometrically, This is a circle of radius $2$ centered at $(1, -2)$ so its only intersection with the $x$-axis is where it's tangent at $(1,0)$. – Ethan Bolker Jan 03 '22 at 01:08
  • Thanks @sean roberson

    I’d like to mark your response as the answer however you replied as a comment so an upvote is the best I can do.

    I see my error.

     – duckegg Jan 03 '22 at 05:56
  • Thanks @hardmath

    I was wondering about how to do that!

    Edited.

     – duckegg Jan 03 '22 at 05:56

1 Answers1

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Plugging in $y=0$ gives me $(x-1)^2+(0+2)^2=4$, from which $(x-1)^2=4-(0+2)^2=0$ and only $x=1$ is obtained. $x=-3$ is nowhere for me.

Oscar Lanzi
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