mmm4444bot
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While determining equations for all lines tangent to the unit circle which pass through the point (4,2), I first found the x-coordinates of the two points of tangency. I will limit my example to the point located in Quadrant IV.
The x-coordinate of this point is (2+?19)/10.
To find the corresponding y-coordinate, I substituted this expression for x into the equation for the unit circle and solved for y. I ended up with the following.
\(\displaystyle \frac{\sqrt{77 - 4 \cdot \sqrt{19}}}{10}\)
Since the larger radicand contains a radical, I will refer to the numerator above as a "compound" radical. I used technology to confirm my result, and I learned that this compound radical simplifies.
\(\displaystyle \frac{1 - 2 \cdot \sqrt{19}}{10}\)
Since it was not immediately obvious to me why the two numerators above are equal, I squared the second one to see how the result compares to the larger radicand in the first one.
\(\displaystyle (1 - 2 \cdot \sqrt{19})^2 \;=\; 1 - 4 \cdot \sqrt{19} + 76\)
This makes it clear why; now, here's my question:
In the future, if I want to use paper and pencil to see if a compound radical similar to this simplifies, how would I proceed?
In hindsight, I see that I could have done the following.
1) Assume that the compound radical takes the form (a + b*?19)^2
2) Write 77 - 4?19 = a[sup:27ic2283]2[/sup:27ic2283] + 2*a*b*?19 + 19*b[sup:27ic2283]2[/sup:27ic2283]
3) By inspection, I need 2*a*b = -4 and a[sup:27ic2283]2[/sup:27ic2283] + 19*b[sup:27ic2283]2[/sup:27ic2283] = 77
4) Solving equations in step 3 simultaneously yields a = ±1 and b = ±2
5) By inspection, a = 1 and b = -2
I did not have the foresight to follow this line of reasoning because I do not know enough to come up with the assumption in step 1.
Cheers,
~ Mark
The x-coordinate of this point is (2+?19)/10.
To find the corresponding y-coordinate, I substituted this expression for x into the equation for the unit circle and solved for y. I ended up with the following.
\(\displaystyle \frac{\sqrt{77 - 4 \cdot \sqrt{19}}}{10}\)
Since the larger radicand contains a radical, I will refer to the numerator above as a "compound" radical. I used technology to confirm my result, and I learned that this compound radical simplifies.
\(\displaystyle \frac{1 - 2 \cdot \sqrt{19}}{10}\)
Since it was not immediately obvious to me why the two numerators above are equal, I squared the second one to see how the result compares to the larger radicand in the first one.
\(\displaystyle (1 - 2 \cdot \sqrt{19})^2 \;=\; 1 - 4 \cdot \sqrt{19} + 76\)
This makes it clear why; now, here's my question:
In the future, if I want to use paper and pencil to see if a compound radical similar to this simplifies, how would I proceed?
In hindsight, I see that I could have done the following.
1) Assume that the compound radical takes the form (a + b*?19)^2
2) Write 77 - 4?19 = a[sup:27ic2283]2[/sup:27ic2283] + 2*a*b*?19 + 19*b[sup:27ic2283]2[/sup:27ic2283]
3) By inspection, I need 2*a*b = -4 and a[sup:27ic2283]2[/sup:27ic2283] + 19*b[sup:27ic2283]2[/sup:27ic2283] = 77
4) Solving equations in step 3 simultaneously yields a = ±1 and b = ±2
5) By inspection, a = 1 and b = -2
I did not have the foresight to follow this line of reasoning because I do not know enough to come up with the assumption in step 1.
Cheers,
~ Mark
