Need help with pythagoras equations

  • Thread starter krislemon
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    Pythagoras
In summary, when solving equations with square roots, one can square both sides and manipulate until no square roots remain. However, this method may introduce extraneous solutions, so it is important to check all solutions by inserting them into the original equation. It is also important to take into account the principal branch of the square root function to avoid confusion with multivalued roots.
  • #1
krislemon
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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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  • #2
krislemon said:
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.
 
  • #3
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.
 
  • #4
krislemon said:
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.
 
  • #5
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.
 
  • #6
Erland said:
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.
 
Last edited:

1. What is the Pythagorean theorem?

The Pythagorean theorem is a mathematical formula that states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

2. How do you use the Pythagorean theorem to find the length of a side?

To find the length of a side in a right triangle using the Pythagorean theorem, you need to know the lengths of the other two sides. Then, you square each of those lengths, add them together, and take the square root of the sum to find the length of the hypotenuse.

3. Can the Pythagorean theorem be used on non-right triangles?

No, the Pythagorean theorem only applies to right triangles. In non-right triangles, the lengths of the sides can be determined using other trigonometric functions such as sine, cosine, and tangent.

4. How can I check if my answer using the Pythagorean theorem is correct?

To check if your answer is correct, you can use a calculator to square the lengths of the other two sides, add them together, and take the square root of the sum. If the result is equal to the length of the hypotenuse, then your answer is correct.

5. Can the Pythagorean theorem be used in real-life applications?

Yes, the Pythagorean theorem has many real-life applications, such as in construction, engineering, and navigation. It is also used in various scientific fields to calculate distances and solve other problems involving right triangles.

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