Answer: c≓−2.135980077
How to solve it: Step by step solution :
STEP
1
:
2c + 5
Simplify ——————
c
Equation at the end of step
1
:
2 (2c+5)
(c-((—•(c2))•(2c-5)))-(((3•(c3))•——————)+2) = 0
6 c
STEP
2
:
Equation at the end of step
2
:
2 (2c+5)
(c-((—•(c2))•(2c-5)))-((3c3•——————)+2) = 0
6 c
STEP
3
:
Dividing exponential expressions
3.1 c3 divided by c1 = c(3 - 1) = c2
Equation at the end of step
3
:
2
(c-((—•(c2))•(2c-5)))-(3c2•(2c+5)+2) = 0
6
STEP
4
:
1
Simplify —
3
Equation at the end of step
4
:
1
(c - ((— • c2) • (2c - 5))) - (6c3 + 15c2 + 2) = 0
3
STEP
5
:
Equation at the end of step 5
c2
(c - (—— • (2c - 5))) - (6c3 + 15c2 + 2) = 0
3
STEP
6
:
Equation at the end of step 6
c2 • (2c - 5)
(c - —————————————) - (6c3 + 15c2 + 2) = 0
3
STEP
7
:
Rewriting the whole as an Equivalent Fraction
7.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using 3 as the denominator :
c c • 3
c = — = —————
1 3
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
7.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
c • 3 - (c2 • (2c-5)) -2c3 + 5c2 + 3c
————————————————————— = ———————————————
3 3
Equation at the end of step
7
:
(-2c3 + 5c2 + 3c)
————————————————— - (6c3 + 15c2 + 2) = 0
3
STEP
8
:
Rewriting the whole as an Equivalent Fraction
8.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 3 as the denominator :
6c3 + 15c2 + 2 (6c3 + 15c2 + 2) • 3
6c3 + 15c2 + 2 = —————————————— = ————————————————————
1 3
STEP
9
:
Pulling out like terms
9.1 Pull out like factors :
-2c3 + 5c2 + 3c = -c • (2c2 - 5c - 3)
Trying to factor by splitting the middle term
9.2 Factoring 2c2 - 5c - 3
The first term is, 2c2 its coefficient is 2 .
The middle term is, -5c its coefficient is -5 .
The last term, "the constant", is -3 Adding up the two equivalent fractions
-c • (c-3) • (2c+1) - ((6c3+15c2+2) • 3) -20c3 - 40c2 + 3c - 6
———————————————————————————————————————— = —————————————————————
3 3
STEP
10
:
Pulling out like terms
10.1 Pull out like factors :
-20c3 - 40c2 + 3c - 6 =
-1 • (20c3 + 40c2 - 3c + 6)
Checking for a perfect cube :
10.2 20c3 + 40c2 - 3c + 6 is not a perfect cube
Solving 2c2-5c-3 = 0 by the Quadratic Formula .
According to the Quadratic Formula, c , the solution for Ac2+Bc+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
c = ————————
2A
In our case, A = 2
B = -5
C = -3
Accordingly, B2 - 4AC =
25 - (-24) =
49
Applying the quadratic formula :
5 ± √ 49
c = —————
4
Can √ 49 be simplified ?
Yes! The prime factorization of 49 is
7•7
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 49 = √ 7•7 =
± 7 • √ 1 =
± 7
So now we are looking at:
c = ( 5 ± 7) / 4
Two real solutions:
c =(5+√49)/4=(5+7)/4= 3.000
or:
c =(5-√49)/4=(5-7)/4= -0.500
One solution was found :
c ≓ -2.135980077
