Can you help with these question
21u^2=u+10
-2x^2 = -(5x+4)
Find the values for c so that s^2 +cs+81=0 has one real root.
^ means to the power of that number that immediately follows
21u^2=u+10
-2x^2 = -(5x+4)
Find the values for c so that s^2 +cs+81=0 has one real root.
^ means to the power of that number that immediately follows
-
Hello
21u^2 = u + 10
u^2 - 1/21*u - 10/21 = 0
u = 1/42 +- √(1/42^2 + 10/21)
u1 = 1/42 + 29/42 = 5/7
u2 = 1/42 - 29/42 = - 2/3
-------------------
- 2x^2 = -5x - 4
x^2 - 5/2*x - 2 = 0
x = 5/4 +-√(25/16 + 2)
x = 5/4 +- √(57/16)
x1 = (5 + √57)/4
x2 = (5 - √57)/4
------------------
s2 +cs + 81 = 0
s = - c/2 +- √(c^2/4 - 81)
the expression under the root must get = 0, so you have only - c/2 as solution:
c^2/4 - 81 = 0
c^2/4 = 81
c^2 = 4*81
c = 18
Regards
21u^2 = u + 10
u^2 - 1/21*u - 10/21 = 0
u = 1/42 +- √(1/42^2 + 10/21)
u1 = 1/42 + 29/42 = 5/7
u2 = 1/42 - 29/42 = - 2/3
-------------------
- 2x^2 = -5x - 4
x^2 - 5/2*x - 2 = 0
x = 5/4 +-√(25/16 + 2)
x = 5/4 +- √(57/16)
x1 = (5 + √57)/4
x2 = (5 - √57)/4
------------------
s2 +cs + 81 = 0
s = - c/2 +- √(c^2/4 - 81)
the expression under the root must get = 0, so you have only - c/2 as solution:
c^2/4 - 81 = 0
c^2/4 = 81
c^2 = 4*81
c = 18
Regards