To figure ut the roots use the quadratic formula
x = [-b +- sqrt(b^2-4ac)]/2a
x = [-k +- sqrt(k^2-4(1)(5)]/2(1)
x = [-k + sqrt(k^2 - 20)]/2 or [-k - sqrt(k^2 - 20)]/2
So the question says these roots differ by sqrt 61, so let's subtract each
[-k + sqrt(k^2 - 20)]/2 - [-k - sqrt(k^2 - 20)]/2
well the k's cancel in the beginning and we are left with 2sqrt(k^2 - 20)/2, and the 2 on top and bottom reduce to
sqrt(k^2 - 20), so this equals sqrt 61
Set equal and solve
sqrt(k^2 - 20) = sqrt(61)
k^2 - 20 = 61
k^2 = 81, so k = +9 or -9
The greatest value therefore is k = +9.
Answer:
The Pythagorean theorem states that a2 + b2 = c2 in a right triangle where c is the longest side. You can use this equation to figure out the length of one side if you have the lengths of the other two.
<span>3.1415926535897 are the first 14 digits</span>
1/4 times 8
=8/4
8 divided by 4 equals
=2