Question

Help w/ Quadratic Equations 2?

Help w/ Quadratic Equations 2?

Hi Can you please solve the following solve for r , please be visual and show your work!

Answer 1

`r = (3+3sqrt3)/2`

Explanation:

Since this triangle has the diameter of the circle as its longest side, it must be a right triangle. Since the radius is `r`, the length of the hypotenuse is `2r`. Now we have the three sides of the triangle and we can solve for `r` using the Pythagorean Theorem:

`(r)^2+(r+3)^2=(2r)^2`

Now we need to expand each of these terms.

`r^2 + (r^2+6r+9) = 4r^2`

And simplify.

`2r^2 + 6r + 9 = 4r^2`

Now, subtract `2r^2 + 6r + 9` from both sides to get a quadratic.

`0 = 2r^2 - 6r - 9`

There are no factors of `-18` which add up to `-6`, so we must use the quadratic formula to solve this.

`r = (-(-6) +- sqrt((-6)^2-4(2)(-9)))/(2(2))`

`r = (6 +- sqrt(36+72))/4 = (6 +- sqrt108)/4`

`r = (6+-6sqrt3)/4 = (3+-3sqrt3)/2`

`r = (3+3sqrt3)/2 " "` or `" " r = (3-3sqrt3)/2`

So we have two solutions. However, the length of the hypotenuse of the triangle cannot be negative. This rules out the second solution.

Therefore, `r = (3+3sqrt3)/2`.

Final Answer

Answer 2

The value of `r=(3+3sqrt3)/2=4.1`

Explanation:

The triangle is a right angle triangle.

So, we can apply Pythagoras ' theorem

`(2r)^2=r^2+(r+3)^2`

`4r^2=r^2+r^2+6r+9`

`4r^2=2r^2+6r+9`

`2r^2-6r-9=0`

The discriminant is

`Delta=b^2-4ac=(-6)^2-4*(2)*(-9)=36+72=108`

As `Delta>0`, there are 2 real solutions

`r=(6+-sqrt(108))/4=(6+-6sqrt3)/4=(3+-3sqrt3)/2`

The 2 solutions are

`r_1=(3+3sqrt3)/2` and `r_2=(3-3sqrt3)/2`

We keep only `r_1` as `r_2<0`