no, you cannot take √ of a negative if you stay in the real number set
So there is no solution.
What are the approximate solutions of 2x^2 -x + 10 = 0
In working this, I came up with a negative (-79) under the radical sign. Can I take the sq rt of a negative?
my choices are:
-2, 2.5
-1.97, 2.47
-2.5, 2
No solution
I think it is no solution. Is this correct? If not, please explain. Thanks.
14 answers
Det x = b^2 - 4ac - formula
x = -b +/- sqrt Det x/2a - formula
if Det x = 0 then x has only 1 solution
Det x > 0 then x has 2 solutions
Det x <0 then x has 2 complex solutions - a+ib=z, z is a complex number, i^2=-i, a is the real part b is the imaginary one
Det x = 1-40=-39
x=1+/-sqrt-39/2=1+/-sqrt i^2*39/2=1+/-i*sqrt39/2
x = -b +/- sqrt Det x/2a - formula
if Det x = 0 then x has only 1 solution
Det x > 0 then x has 2 solutions
Det x <0 then x has 2 complex solutions - a+ib=z, z is a complex number, i^2=-i, a is the real part b is the imaginary one
Det x = 1-40=-39
x=1+/-sqrt-39/2=1+/-sqrt i^2*39/2=1+/-i*sqrt39/2
sorry there was a miscalculation.
Det x = 1-80=-79 not -39
Det x = 1-80=-79 not -39
also i^2=-1
For Connexus Students (Alg.1 Unit 5 Lesson 6 practice test)
1. B
2. C
3. D
4. A
5. D
6. D
7. C
8. A
1. B
2. C
3. D
4. A
5. D
6. D
7. C
8. A
posted by is 100% correct!! Thx!!
Hes actually right
2021 posted by is still right
thanks posted by i got a 100
feb 10th still works!
They are actually good,as of March 2nd 2022.
Thx guys
They are actually good,as of March 2nd 2023.
3/3/23 still works got 100% thx
1 answer