1 answer:
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.
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The element gold (au) occurs in the 6th period
Answer:
a=70 vertically opp angles
b=80 linear pair
c=100 corresponding angles
d= 80 linear pair with c
e= 180 -a-d
=180-70-80
=30
Answer:
2
Step-by-step explanation:
The slope is given by
m = ( y2-y1)/(x2-x1)
= (7-3)/(10-8)
= 4/2
= 2
113/12 into a mix number is 9 5/12.
M=-1/3
that is the answer