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Need help with pythagoras equations

by krislemon
Tags: equations, pythagoras
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krislemon
#1
Nov22-12, 05:56 AM
P: 2
Hi,

I need help with an explanation of how to deal with squareroots when solving equations.
E.g. Look at the uploaded bmp.file and solve for x. Please I would like to have a full step by step method on how to deal with this type of equations in general. Thank you
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File Type: bmp squareroot equation.bmp (45.4 KB, 28 views)
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Mentallic
#2
Nov22-12, 06:14 AM
HW Helper
P: 3,535
Quote Quote by krislemon View Post
Hi,

I need help with an explanation of how to deal with squareroots when solving equations.
E.g. Look at the uploaded bmp.file and solve for x. Please I would like to have a full step by step method on how to deal with this type of equations in general. Thank you
Sorry but it's forum rules that we can't just give you a full step by step solution. You need to show you've put some effort into the problem yourself.

Start by squaring both sides, then show us what you've done.
krislemon
#3
Nov22-12, 06:22 AM
P: 2
ok.
I managed to solve the problem so I will just ask about the squareroot.
Where can I find the theory/rule of canceling squareroot? e.g. (squareroot of x)^2.

Mentallic
#4
Nov22-12, 08:36 AM
HW Helper
P: 3,535
Need help with pythagoras equations

Quote Quote by krislemon View Post
Where can I find the theory/rule of canceling squareroot? e.g. (squareroot of x)^2.
I'm not sure what you're asking exactly. The theory of cancelling the square root? Squaring is the inverse of the square root (cancelling the square root) as you've shown.

If you want to read a bit more on square roots, maybe try wikipedia? Or you'll need to be more specific.
Erland
#5
Nov23-12, 08:04 AM
P: 345
When solving problems of this kind, one can square both sides and manipulate until no square roots remain. But when squaring, the implication goes only forward. Therefore, one must check all solutions one finds by inserting them into the original equation.

For example, squaring the equation ##\sqrt x =-1## gives ##x=1##. The first equation can therefore have no solution other than ##x=1##. But if we insert this in the first eqaution, we obtain ##\sqrt 1=-1##, which is false. Therefore, ##x=1## is not a solution either, so the first equation has no solutions.
Millennial
#6
Nov23-12, 08:11 AM
P: 295
Quote Quote by Erland View Post
When solving problems of this kind, one can square both sides and manipulate until no square roots remain. But when squaring, the implication goes only forward. Therefore, one must check all solutions one finds by inserting them into the original equation.

For example, squaring the equation ##\sqrt x =-1## gives ##x=1##. The first equation can therefore have no solution other than ##x=1##. But if we insert this in the first eqaution, we obtain ##\sqrt 1=-1##, which is false. Therefore, ##x=1## is not a solution either, so the first equation has no solutions.
I hope you took the square root function in the sense of the principal branch, for if you did not, we have that [itex]\sqrt{}[/itex] is a multivalued function, having precisely two values for every real number other than 0, and it is one-valued for 0. Taking the square root as multivalued yields [itex]-1\in \sqrt{1}[/itex], which is true, for [itex](-1)^2=1[/itex]. However, the square root in the principal branch is defined as the positive answer to the multivalued square root, making it one-valued. In general, [itex]\sqrt[n]{}[/itex] is a multivalued function, having n values for all complex numbers and 1 for 0. For example, the real number 2 has three cube roots: [itex]\sqrt[3]{2}[/itex] which is the positive root, and two complex roots: [itex]\displaystyle -\frac{\sqrt[3]{2}}{2}+\frac{\sqrt[3]{2}\sqrt{3}}{2}i[/itex] and [itex]\displaystyle -\frac{\sqrt[3]{2}}{2}-\frac{\sqrt[3]{2}\sqrt{3}}{2}i[/itex]. We avoid this problem by defining the result of a rooting operation to satisfy these if x is real:

[itex]\sqrt[n]{x}[/itex]'s principal branch value is defined so as to be the unique real number a satisfying [itex]a\in\sqrt[n_m]{x}[/itex] and [itex]a\geq 0[/itex], where [itex]\sqrt[n_m]{}[/itex] denotes the multivalued version of the nth root.

I just wanted to point that out to avoid confusion.


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