mathematics?

Question

1)Find the value of c that makes this trinomial a perfect square.
x2 + 17x + c

if u know how to answer this... please explain

2) Write this expression in the standard form a + bi.
a) 9i(2 - 3i) b) (3 - 4i)(2 + i) c) 10 / (3-4i)

i have no idea what the 'i' means, please explain what 'i' is if possible.

3) Find the exact solution for this equation by completing the square.
2x2 - 7x + 12 = 0
Answer the question using the form x = [A +/- Bi (square root) C] / D where A is positive.

i think if i understand what 'i' is, i might be able to do this.. but just to make sure.

Reply to Skittless qusetion... :
yes these are questions out of 149 questions, out of a holiday homework assignment... these are the ones i am stuck on. if i just wanted to finish my homework i would not have asked for an explanation for each answer.

PLEASE READ THIS =D

thanks to all of the answers i got, i now understand everything.
But i just cant figure this out.

Perform the indicated operations and write the answer in standard form.

(6-4i)/(3-i)

Give your answer using the form Ai + B.

can anyone explain how to do this?

thanks

Answer 1

1) to make this trinomial a perfect square, you'd need
x^2 + 17x + (sqrt c)^2 where 2 * 1 * sqrt c = 17
This is because in general (ax + b)^2 = (ax)^2 + (2abx) + b^2. Since a = 1 and b^2 = c, b = sqrt c.
So, solving for c, you'd get sqrt c = (17/2) so c = (17/2)^2
Not a nice number but there you go.

2) i is the number whose square is -1. (Technically, it's not the square root of -1, because every number has two square roots, positivs and -, and this gets tricky when you're talking about i)
So to do this, just multiply it out like you ordinarily do binomials, only whenever you encounter i^2, replace it with (-1)
So you have ... 9i(2-3i) = 18i - 27i^2 = 18i - (-27) = 18i + 27
Then we usually rearrange the numbers (since addition is commutative) so that the "i" part is second, and get
27 + 18i
For the next one, multiply it out like binomials, (FOIL) and agian, whenver you encounter an i^2, replace it with (-1)...
(3-4i)(2+i) = 6 + 3i - 8i -4(i^2) =
6 - 5i -(-4) = 6 - 5i + 4 = 10 - 5i
For the next one, rationalize the denominator by multiplying top and bottom by the conjugate of the denominator...
(10/(3-4i)) * (3+4i)/(3+4i) The reason this works is because the product of two conjugates is the difference of two sqaures, and, since you have an i^2 in there it will work out to the SUM of two squares....
10(3 + 4i)/((3+4i)(3-4i)) = 10(3+4i)/(3^2 -((4i)^2)) = (30+40i)/(9-(-16)) = (30+40i)/ 25 Notice how your denominator should always come out to the sum of two square real numbers which is again a real number.
3) No need for me to work this one out, as others have done this just fine. But I suggest you check out some of the online tutoring sites for a better explanation of these concepts. The graphics are much better so the explanations would be easier to follow.
http://amby.com/educate/math
http://www.math.com
http://www.webmath.com

PS -- Jacquelyn, below, copied and pasted my answer! No fair!

Answer 2

1) We're looking for the value of c so that the expression can be rewritten as something squared. Since we're dealing with a trinomial, the expression should end up looking like (p*x + q)^2, where p and q are some constants. If you multiply that out (using something like the F.O.I.L. method), you get

(p*x + q) * (p*x + q)
p^2*x^2 + 2pqx + q^2

We want this to match up with x^2 + 17x + c. This means the coefficients have to match up. So look at this term by term:

p^2*x^2 = x^2
2(pq)x = 17x
q^2 = c

From the first term, we see that p has to equal 1 (it could also be -1, but let's ignore that for now). Plug p into the second term, and you get 2(1*q)x = 17x

Divide both sides by x and solve for q:

2(q) = 17
q = (17/2)

We said q^2 = c, so this means c equals 17/2 squared, or 289/4. That's our answer: c=289/4. Note that when c=289/4, then x^2 + 17x + 25/4 = (x + 17/2) ^ 2, a perfect square.

Earlier I said to ignore the possibility that p= -1. That's because if you work it out, it will just give you q = -17/2, which squared gives you the same answer. This has to do with the fact that (-x + -17/2)^2 gives you the original equation too.

2) "i" is the constant representing the square root of -1. This is the most fundamental concept of imaginary and complex numbers and how they work. Since you're not familiar with it, you should research it and how it works. Are you working from a text book?

3) Completeing-the-square is a method that involves getting the trinomial into a perfect square on one side. Go here for an explanation: http://mathforum.org/library/drmath/view/53165.html
Once you have it in proper form, take the square root of both sides (making use of "i") and solve the equation for "x".

Answer 3

Here is your answer And YES Iam smart it takes while while but it took me five minute to copy and paste off my worksheet. So here you go

1) to make this trinomial a perfect square, you'd need
x^2 + 17x + (sqrt c)^2 where 2 * 1 * sqrt c = 17
This is because in general (ax + b)^2 = (ax)^2 + (2abx) + b^2. Since a = 1 and b^2 = c, b = sqrt c.
So, solving for c, you'd get sqrt c = (17/2) so c = (17/2)^2
Not a nice number but there you go.

2) i is the number whose square is -1. (Technically, it's not the square root of -1, because every number has two square roots, positivs and -, and this gets tricky when you're talking about i)
So to do this, just multiply it out like you ordinarily do binomials, only whenever you encounter i^2, replace it with (-1)
So you have ... 9i(2-3i) = 18i - 27i^2 = 18i - (-27) = 18i + 27
Then we usually rearrange the numbers (since addition is commutative) so that the "i" part is second, and get
27 + 18i
For the next one, multiply it out like binomials, (FOIL) and agian, whenver you encounter an i^2, replace it with (-1)...
(3-4i)(2+i) = 6 + 3i - 8i -4(i^2) =
6 - 5i -(-4) = 6 - 5i + 4 = 10 - 5i
For the next one, rationalize the denominator by multiplying top and bottom by the conjugate of the denominator...
(10/(3-4i)) * (3+4i)/(3+4i) The reason this works is because the product of two conjugates is the difference of two sqaures, and, since you have an i^2 in there it will work out to the SUM of two squares....
10(3 + 4i)/((3+4i)(3-4i)) = 10(3+4i)/(3^2 -((4i)^2)) = (30+40i)/(9-(-16)) = (30+40i)/ 25 Notice how your denominator should always come out to the sum of two square real numbers which is again a real number.
3) No need for me to work this one out, as others have done this just fine. But I suggest you check out some of the online tutoring sites for a better explanation of these concepts. The graphics are much better so the explanations would be easier to follow.
http://amby.com/educate/math
http://www.math.com
http://www.webmath.com

Answer 4

1) let x^2+17x+c=(ax+b)^2
= a^2x^2+2abx+b^2

therfore a^2x^2= x^2
2abx = 17x
and
b^2 = c

so a=1 and 2bx = 17x
so b = 17/2
so c=(17/2)^2 = 289/4

2)
i = sqrt(-1) it's called an imaginary number

a) distribute the 9i
18i-27*(i)^2
since i = sqrt(-1), i^2 = -1
18i+27 or 27+18i

b) is basically the same thing

c) just like rationalizing the denominator using a conjugate, where the denominator is a+bi you mutliply by (a-bi)/(a-bi) which is = 1

10(3+4i)/((3-4i)(3+4i))
(30+40i)/(9-12i+12i-16(i)^2)
(30+40i)/(9+16)
(30+40i)/25 or 30/25+40/25 i = 6/5+ 8/5 i

3) make the coefficient of x^2 = 1 and move the contant to the other side
x^2-7/2x=-6
if the coefficient of x is a then add (a/2)^2 to both sides
so ((-7/2)/2)^2 which is (-7/4)^2 = 49 /16
x^2-7/2x+49/16= -6+49/16 = -47/16
now the left is a perfect square
(x-7/4)^2 = -47/16
sqrt both sides
x-7/4 = +/- sqrt(-47/16)
= +/- sqrt(-1* 47/16) = +/- i sqrt(47)/4
and add 7/4 to both sides
x = 7/4 +/- i sqrt(47)/4
= (7 +/- i sqrt(47)/4

Answer 5

The first two have been solved. Here I'd like to try the third.

3) Find the exact solution for this equation by completing the square.
2x^2 - 7x + 12 = 0

Divide by 2 and move the constant term to the other side,
x^2 - (7/2)x = -6

Add (7/4)^2 on both sides and complete the square,
(x - 7/4)^2 = -6 + (7/4)^2

Solve for x by taking square root, moving the contant term to the right side, and transforming to the required form,

x = 7/4 ± i√[6 - (7/4)^2] = [7 ± i√47]/4

Answer 6

1) x^2 + 17x + c
= (x + 17/2)(x + 17/2)
= x^2 + 17x + 72.25


2) i = sqrt(-1)
i^2 = -1

a) 9i(2 - 3i)
= 18i - 3i^2
= 3 + 18i

b) (3 - 4i)(2 + i)
= 6 + 3i - 8i - 4i^2
= 10 - 5i

c) 10 / (3 - 4i)
= [ 10 / (3 - 4i) ] * [ (3 + 4i) / (3 + 4i) ]
= 10(3 + 4i) / (9 - 16i^2)
= (30 + 40i) / 25
= 6/5 + (8/5)i


3)
2x2 - 7x + 12 = 0
2x2 - 7x = -12
x2 - 7/2 = -6
x2 - 7/2 + 49/4 = -6 + 49/4
(x - 7/4)^2 = 25/4
x - 7/4 = +/- sqrt(25/4)
x = 7/4 + 5/2 x = 7/4 - 5/2
x = 17/4 x = - 3/4

Answer 7

1)first find D for the eq
a=1,b=17,c=c
now d=b^2-4ac
since it is perfect ^2 therefore d=0
17^2-4*1*c=0
289-4c=0
-4c=-289
c=289/4
2)i=iota ie(-1^1/2)remember i^2=-1
a)18i+27
b)6+3i-8i+4
-5I=10
c)divide&mutliply it by(3+4i)
then u will get
30+40i/9+16
30+40i/25
6/5+8/5i

Answer 8

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Answer 9

I know #3 now. The others take too long.

3) 2x 2-7x+12=0
12+2+2x+2-7x=0+2
14+2x+2-7x=2
14+2x*-7x+2=2*-7x
-14x+2=2*-7x
-14x+2-2=2-2*-7x
-14x=-7x
14x/14=7x/14




I hope I got it!

Answer 10

i is the square root of -1. So i squared = -1