Question

I have done this question countless times and I have been wrong can someone tell me where I went wrong on this......

Answer 1

Thank you

Answer 2

can you post a screen shot? i have a different Microsoft word and can't open the file :/

Answer 3

How do you do that

Answer 4

Do you still need help? Or have you figured it out?

Answer 5

do you have a mac or a pc? i dont know how to do it on a pc but i assume you can just google it :D

Answer 6

then you just attach the file the same way you attached your document

Answer 7

Problem 1) 50% correct; repost
Solve by factoring: 2a2 − 3a = −5
Solution
Your work
2a^2-3a=-5
2a^2-3a+5=0
a=(-b±√(b^2-4ac))/2a
aa^2+ba+c=0
a=2, b=-3, c=5
a=(-(-3)±√(〖(-3)〗^2-4(2)(5)))/(2(2))
a=(3±√((〖-3)〗^2-4(2)(5)))/(2(2))
a=(3±i√31)/(2(2))

a=(3±i√31)/4

a=(3+i√31)/4

a=(3-i√31)/4
A=(3+i√31)/4, (3-i√31)/4

My comments
You are close, but you need to do some more work.
After showing the following equation
a=(3±√((〖-3)〗^2-4(2)(5)))/(2(2))
What is the sign of the discriminant ?
What conclusions can you make base on this sign ?
Your result below is a complex number, but you need to see if there is a solution in the set of real numbers. This result is correct in general, but the problem is expecting a solution that is a real number.
a=(3±i√31)/4

Hint:
Find the discriminant, and depending on the sign, find if the solution exist or not, and what type of polynomial you have.
The discriminant is b^2-4ac, where the equation is of the form a^2 + bx + c.
Since you have 2x^2-3x+5=0 (I replaced a by x to avoid confusion)
Then, a=2, b=-3 and c=5.

Answer 8

^^^^^^^^
From file

Answer 9

Thats it thank you zordo

Answer 10

9 minus 40 is negative 31

Answer 11

Take a look at this:

Answer 12

But where did I go wrong on the answer?