circle problem

[apple2357]

This is a little odd to me.
I can find a way of getting x using some geometry using right angled triangles but can't see how to get to this quadratic equation.
Any ideas?

 

[Dr.Peterson]

This is a little odd to me.
I can find a way of getting x using some geometry using right angled triangles but can't see how to get to this quadratic equation.
Any ideas?

Write an equation that adds some lengths to equal 10 (it will involve a radical), and arrange it so that the radical appears on one side, then square both sides and rearrange.

Once you've shown some equation, it will be easier to give hints without giving the whole thing away.

Of course, you don't need that equation to solve the problem, so it's a little contrived -- I don't care for problems that say "do it my way, even though it's not the best, and I won't tell you what my way is".

 

[apple2357]

Write an equation that adds some lengths to equal 10 (it will involve a radical), and arrange it so that the radical appears on one side, then square both sides and rearrange.

Once you've shown some equation, it will be easier to give hints without giving the whole thing away.

Of course, you don't need that equation to solve the problem, so it's a little contrived -- I don't care for problems that say "do it my way, even though it's not the best, and I won't tell you what my way is".

Ok i think i have got it. I am looking at a triangle where the two circles touch.

And i seem to get (x+4)^2 = (6-x)^2 + (6-x)^2
This looks like it give the quadratic.

The other approach i can think of is thinking about the length of the diagonal:


x + x root 2 + 4 + 4 root 2 = 10 root 2

But this only gives me one solution and not the quadratic

 

[pka]

Ok i think i have got it. I am looking at a triangle where the two circles touch.
And i seem to get (x+4)^2 = (6-x)^2 + (6-x)^2
This looks like it give the quadratic.

HINT:
\(\displaystyle \begin{align*}(x+4)^2&=2(x-6)^2 \\x^2+8x+16&=2(36-12x+x^2)\\x^2+8x+16&=72-24x+2x^2\\0&=~? \end{align*}\)

 

[Dr.Peterson]

Ok i think i have got it. I am looking at a triangle where the two circles touch.

And i seem to get (x+4)^2 = (6-x)^2 + (6-x)^2
This looks like it give the quadratic.

Yes, as pka showed, this leads to the desired quadratic.

The other approach i can think of is thinking about the length of the diagonal:

x + x root 2 + 4 + 4 root 2 = 10 root 2

But this only gives me one solution and not the quadratic

You can in fact get from here to the quadratic as well; this is closer to what I described. We can gather the radical terms on the right:

x + x sqrt(2) + 4 + 4 sqrt(2) = 10 sqrt(2)

x + 4 = 10 sqrt(2) - 4 sqrt(2) - x sqrt(2)

x + 4 = 6 sqrt(2) - x sqrt(2)

x + 4 = (6 - x)sqrt(2)

Square both sides, and you get the same equation, which you can expand and simplify to the required form.

 

[Steven G]

x + x root 2 + 4 + 4 root 2 = 10 root 2

But this only gives me one solution and not the quadratic

But will both solutions work for this problem. Suppose for example in finding the length s of a side of a rectangle yields that s² = 100. Then s= +/- 10. S=-10 makes no sense, as the side of a rectangle can't be negative.

Actually your equation, x + x root 2 + 4 + 4 root 2 = 10 root 2, yields the correct (and only) answer. As Dr P pointed out, if you square both sides you get the quadratic that you desired. However squaring can give invalid results. Consider that x=-2, so x equals -2 and nothing else as this is given. Now if you square both sides you get x² = 4. Then x= +/- 2. Clearly x=2 is not a valid solution! Be careful with squaring!

 

[Dr.Peterson]

Actually your equation, x + x root 2 + 4 + 4 root 2 = 10 root 2, yields the correct (and only) answer. As Dr P pointed out, if you square both sides you get the quadratic that you desired. However squaring can give invalid results. Consider that x=-2, so x equals -2 and nothing else as this is given. Now if you square both sides you get x² = 4. Then x= +/- 2. Clearly x=2 is not a valid solution! Be careful with squaring!

An excellent point: The known solution is a solution of the quadratic, but so is an additional, negative solution, which we call "extraneous". All you were asked to do is to show that the quadratic is satisfied by the solution, not that all solutions of the quadratic are valid solutions of the problem, which is not true.

No one has said that finding the quadratic is a better solution method than what you had already done ...

 

[Denis]

Then x= +/- 2. Clearly x=2 is not a valid solution! Be careful with squaring!

Sad sack was sittin' on a block of stone
Way over in the corner weepin' all alone
The warden said, "hey, buddy, don't you be no square <<<<<<<<<
If you can't find a partner, use a wooden chair"