Question

3 Parts to this question
a. Find the value of the discriminant
b. Give the nukber of real solutions
c. Find the real solutions, rounded to the nearest hundred.

r^(2) + 6r + 4 = 0

(this is a reallu weird question so if you dont understand it i understand)

Answer 1

The discriminant = \[b ^{2}-4ac = 6^{2}-4*1*4 = 36 - 32 = 4\]

# of Real Solutions: 2

Real solutions using quadratic formula (check my algebra carefully cause I tend to get sloppy):
\[(-b \pm \sqrt{b ^{2}-4ac})/2a\]

\[(-(-6) \pm \sqrt{6^{2}-4*1*4)}/2*1\]

= \[6 \pm \sqrt{36-32}/2=6 \pm \sqrt {4}/2 = 3 \pm 1 = 4,2\]

Answer 2

And check my work via FOIL or whatever method you prefer.

Answer 3

Oh, and something is wrong since 4*2 does not equal 4.

Answer 4

im confused how did ur 6 become -(-6) when it was positive to begin with shouldnt it have become -(6) and htere for -6 + or - not 6 + or -?

Answer 5

Yep.

Answer 6

Like I said I get sloppy with my algebra and arithmetic. Sorry. I think something is still wrong in there somewhere but I'm going to be lazy with it.

Answer 7

atleast it was caught XD