Asked by Jane

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.
Answered by Reiny
no, you cannot take √ of a negative if you stay in the real number set
So there is no solution.
Answered by Chris
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
Answered by Chris
sorry there was a miscalculation.
Det x = 1-80=-79 not -39
Answered by Chris
also i^2=-1
Answered by posted by
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
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