Thursday

March 5, 2015

March 5, 2015

Posted by **Shay** on Thursday, January 18, 2007 at 4:53pm.

(3+2i)-(-7-i)

10 + 3i

how did you do that?

can you show me this one and explain how this time?

(3+2i)(-7-i)

or even

(3-2i)^2

you would do 3+2i=5i, then -7-i=7i then 5i+7i=12i

(3+2i)-(-7-i) =

(3+2i) + (7 + i) =

3 + 7 + 2i + i =

10 + 3i

(3+2i)*(-7-i) =

-(3+2i)*(7 + i) =

-[3*7 + 3*i + 2i*7 +2i*i] =

-[21 + 3i + 14i - 2] =

-[19 + 17i]

(3-2i)^2 =

9 - 12i - 4 =

5 - 12 i

thankyou, you are very good at math!

can you help with this one please and thankyou?

Determine the values of k if the graph of

y=2x^2-sx+3k

intersects the x-axis at 2 distinct points.

Thanks!

a x^2 + bx + c = 0 has two different roots if the discriminant D defined as:

D = b^2 - 4 a c does not equal zero.

If a, b and c are real numbers then for D > 0 there are two real roots while for D < 0 there are two complex roots.

In this problem we have to demand that D > 0 --->

D = s^2 - 24 k > 0 --->

k < s^2/24

- can someone please help me? -
**hi**, Wednesday, February 3, 2010 at 9:27pmif anyone knows the answer please help me on this one too!

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