five over c plus 2 minus 5 over c squared minus 4 equal 0. solve for c?

Answer 1

5/(c + 2) - 5/(c^2 - 4) = 0

Hopefully this is the equation of your problem.

c^2 - 4 is a difference of two squares and factors into (c + 2)(c - 2)

The LCD of the fractions is (c + 2)(c - 2), and we can clear the fractions by multiplying both sides of the equation by the LCD.

5(c - 2) - 5 = 0
5c - 10 - 5 = 0
5c = 15
c = 3

Check
5/(c + 2) - 5/(c^2 - 4) = 0
5/(3 + 2) - 5/(9 - 4) = 0
5/5 - 5/5 = 0
0 = 0

Answer 2

I see your equation as this: 5/c + (2-5)/c^2 -4=0
5/c + (-3)/c^2 -4=0 Do brackets first
5/c +(-3)/c^2 -4=0+4 Get rid of the 4; add 4
5/c +(-3)/c^2 =4
5/c +(-3)/c^2=4x5 Get rid of the 5; multiply by 5
c +(-3)/c^2=20
c+ (-3)/c^2=20x(-3) Get rid of the (-3)
c+c^2=(-60)
c^3=(-60) Use the product rule to collect like terms
3√c^3=3√(-60) Cubic root gets rid of the ^3
c= -3.9 (approx)

Answer 3

I'm guessing where the parentheses are:

5/(c+2) - 5/(c^2) - 4 = 0
Am I right? Hope not, because if I am, we get a cubic out of it!

Mult each term by c^2*(c+2):
5c^2 - 5(c+2) - 4c^2*(c+2) = 0
Mult by -1 and remove parentheses:
4c^3 + 3c^2 + 5c + 10 = 0.

I'm sure there are no rational solutions, but I think there's an irrational one between -1 and -2!!

Answer 4

Yes you will be solving for c but c will have 2 answers because you will be using the quadratic equation:

c= -b + or - sqr(b^2 - 4ac)/2a

But first we have to get your equation in the form:
ac^2 + bc + c = 0

As I understand your problem you have: 5/c + 2 - (5/c)^2 - 4 = 0

Step 1: -5/c * -5/c = 25/c^2 rewrite equation
5/c + 2 + 25/c^2 - 4 = 0

Step 2: add 4 to both sides of the equation and subtact 2 from both side os the equation and rewrite.
5/c + 25/c^2 = 2

Step 3: Multiply both sides of the equation by c^2 and rewrite.
5/c + 25 = 2c^2

Step 4: subtract 25 from both sides and rewrite
5/c = 2c^2 - 25

Step 5: Multiply both sides by c and rewrite
5 = 2c^2 - 25c

Step 6: subtract 5 from both sides and rewrite
2c^2 - 25c - 5 = 0 almost there.

Step 7: multiply all terms by -1 to get correct form and rewrite
-2c^2 + 25c + 5 = 0

Step 8: Quadratic equation time
c = -b + sqr(b^2 - 4ac)/2a and c = -b - sqr(b^2 - 4ac)/2a

Plug in your variables and solve and as always "Please Excuse My Dear Aunt Sally"

Answer 5

ok. So, you want to place in writing a worry-loose equation for a parabola from a graph. the style will look as though this: y=Ax^2+Bx+C yet, you ought to prepare a diverse style; in all probability Vertex style. It feels like this: y=a(x-h)^2+ok because of the fact you have the vertex already (4,7), it is your (h,ok). so as which you will plug that into the equation. y=a(x-4)^2+7 so a techniques as getting a, i'm now not powerful how, yet you ought to redecorate Vertex style into primary style. superb that's an social amassing on the because of the fact of attempt this, for y=-2(x-2)^2+3: y=-2(x-2)^2+3 y=-2 [(x-2)(x-2)]+3 y=-2(x^2-4x+4)+3 y=-2x^2+8x-8+3 y=-2x^2+8x-5 notice: Now, you stated it opens down, so this is going to permit you to acknowledge that A is destructive, or a<0. want this supplies an thought on the because of the fact of do it. you have gotten to determine on the innovations on your individual.

Answer 6

you sound like my math teacher mr clark

Answer 7

maybe you should put it in an equation.

Answer 8

i dont see a solution.