Question 13 (Multiple Choice Worth 5 points) [9.05] What are the exact solutions of x2 − x − 4 = 0? x = the quantity of negative 1 plus or minus the square root of 15 all over 2 x = the quantity of 1 plus or minus the square root of 15 all over 2 x = the quantity of 1 plus or minus the square root of 17 all over 2 x = the quantity of negative 1 plus or minus the square root of 17 all over 2
lol ;)
I think it be D but its not right to me can you help me
so, lets use the difference of squares method (also know as the pathagorian therom) to solve this.
\[(-b +or- \sqrt{b^2-4*ac})/2a\]
whats next
so the equation us have has the form a^2-b-c=0
sooo....\[-1+or-(\sqrt{-1^2-4*1*-4})/2*1\]
I don't know how to do this sorry
solve for -1^2-4*1*-4 first...k
I am lost sorry
okay sorry... what does -1^2*4*1*-4=?
opps I meant -1^2-4*1*-4=?
how do i solve for that
-1^2=1 right-4*1which is -4*-4 which =16...so ur left with 17 right?
yea
-1^2 means negative one to the second power. you know that ya?
2 right
-1x -1 = -1^2 = 1
oh ok
k
so the answer is C right
YOUUUUUU GOT IT... NICE JOB.
cool one more thing can you help me on this Question 14 (Multiple Choice Worth 5 points) [9.05] What are the approximate solutions of 2x2 − 7x = 3 rounded to the nearest hundredth? No real solutions x ≈ −0.77 and x ≈ 7.77 x ≈ −0.39 and x ≈ 3.89 x ≈ −3.89 and x ≈ 0.39
I think its A but not sure lol
k..use the pathagorian therom to solve this one too..but u will need a calc for this one
the 3 on the right side of the equation is confusing u hu?
yea
will it B
so just subtract 3 from both sides so that the right side is equal to 0... like this \[2x^2-7x-3=3-3 -> 2x^2-7x-3=0\] see,
now put it in your calc \[-1*-7 +or- (\sqrt{-7^2-4*2*-3})/2*2\]
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