Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
((v-6)/(v+2)+1)-((v-2)/(v+4))=0
Step by step solution :
Step 1 :
v - 2 Simplify ————— v + 4Equation at the end of step 1 :
(v - 6) (v - 2) (——————— + 1) - ——————— = 0 (v + 2) v + 4 Step 2 :
v - 6 Simplify ————— v + 2Equation at the end of step 2 :
(v - 6) (v - 2) (——————— + 1) - ——————— = 0 v + 2 v + 4 Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1Adding a whole to a fraction
Rewrite the whole as a fraction using (v+2) as the denominator :
1 1 • (v + 2) 1 = — = ——————————— 1 (v + 2) 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 :
3.2 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:
(v-6) + 1 • v+2 2v - 4 ——————————————— = ——————————— 1 • (v+2) 1 • (v + 2)Equation at the end of step 3 :
(2v - 4) (v - 2) ——————————— - ——————— = 0 1 • (v + 2) v + 4 Step 4 :
Step 5 :
Pulling out like terms :
5.1 Pull out like factors:
2v - 4=2•(v - 2)
Calculating the Least Common Multiple :
5.2 Find the Least Common Multiple
The left denominator is : v + 2
The right denominator is : v + 4
| Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| v + 2 | 1 | 0 | 1 |
| v + 4 | 0 | 1 | 1 |
Least Common Multiple:
(v + 2)•(v + 4)
Calculating Multipliers :
5.3 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M=L.C.M/L_Deno=v + 4
Right_M=L.C.M/R_Deno=v + 2
Making Equivalent Fractions :
5.4 Rewrite the two fractions into equivalent fractions
Two fractions are called equivalent if they have the same numeric value.
For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.
To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.
L. Mult. • L. Num. 2 • (v-2) • (v+4) —————————————————— = ————————————————— L.C.M (v+2) • (v+4) R. Mult. • R. Num. (v-2) • (v+2) —————————————————— = ————————————— L.C.M (v+2) • (v+4)
Adding fractions that have a common denominator :
5.5 Adding up the two equivalent fractions
2 • (v-2) • (v+4) - ((v-2) • (v+2)) v2 + 4v - 12 ——————————————————————————————————— = ————————————————— (v+2) • (v+4) (v + 2) • (v + 4)Trying to factor by splitting the middle term
5.6Factoring v2 + 4v - 12
The first term is, v2 its coefficient is 1.
The middle term is, +4v its coefficient is 4.
The last term, "the constant", is -12
Step-1 : Multiply the coefficient of the first term by the constant 1•-12=-12
Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is 4.
| -12 | + | 1 | = | -11 | ||
| -6 | + | 2 | = | -4 | ||
| -4 | + | 3 | = | -1 | ||
| -3 | + | 4 | = | 1 | ||
| -2 | + | 6 | = | 4 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -2 and 6
v2 - 2v+6v - 12
Step-4 : Add up the first 2 terms, pulling out like factors:
v•(v-2)
Add up the last 2 terms, pulling out common factors:
6•(v-2)
Step-5:Add up the four terms of step4:
(v+6)•(v-2)
Which is the desired factorization
Equation at the end of step 5 :
(v + 6) • (v - 2) ————————————————— = 0 (v + 2) • (v + 4)Step 6 :
When a fraction equals zero :
6.1 When a fraction equals zero ...Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
(v+6)•(v-2) ——————————— • (v+2)•(v+4) = 0 • (v+2)•(v+4) (v+2)•(v+4)Now, on the left hand side, the (v+2)•(v+4) cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape:
(v+6) • (v-2)=0
Theory - Roots of a product :
6.2 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation:
6.3Solve:v+6 = 0Subtract 6 from both sides of the equation:
v = -6
Solving a Single Variable Equation:
6.4Solve:v-2 = 0Add 2 to both sides of the equation:
v = 2
Supplement : Solving Quadratic Equation Directly
Solving v2+4v-12 = 0 directly Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex:
7.1Find the Vertex ofy = v2+4v-12Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,1, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Av2+Bv+C,the v-coordinate of the vertex is given by -B/(2A). In our case the v coordinate is -2.0000Plugging into the parabola formula -2.0000 for v we can calculate the y-coordinate:
y = 1.0 * -2.00 * -2.00 + 4.0 * -2.00 - 12.0
or y = -16.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = v2+4v-12
Axis of Symmetry (dashed) {v}={-2.00}
Vertex at {v,y} = {-2.00,-16.00}
v-Intercepts (Roots) :
Root 1 at {v,y} = {-6.00, 0.00}
Root 2 at {v,y} = { 2.00, 0.00}
Solve Quadratic Equation by Completing The Square
7.2Solvingv2+4v-12 = 0 by Completing The Square.Add 12 to both side of the equation :
v2+4v = 12
Now the clever bit: Take the coefficient of v, which is 4, divide by two, giving 2, and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have:
12+4or, (12/1)+(4/1)
The common denominator of the two fractions is 1Adding (12/1)+(4/1) gives 16/1
So adding to both sides we finally get:
v2+4v+4 = 16
Adding 4 has completed the left hand side into a perfect square :
v2+4v+4=
(v+2)•(v+2)=
(v+2)2
Things which are equal to the same thing are also equal to one another. Since
v2+4v+4 = 16 and
v2+4v+4 = (v+2)2
then, according to the law of transitivity,
(v+2)2 = 16
We'll refer to this Equation as Eq. #7.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(v+2)2 is
(v+2)2/2=
(v+2)1=
v+2
Now, applying the Square Root Principle to Eq.#7.2.1 we get:
v+2= √ 16
Subtract 2 from both sides to obtain:
v = -2 + √ 16
Since a square root has two values, one positive and the other negative
v2 + 4v - 12 = 0
has two solutions:
v = -2 + √ 16
or
v = -2 - √ 16
Solve Quadratic Equation using the Quadratic Formula
7.3Solvingv2+4v-12 = 0 by the Quadratic Formula.According to the Quadratic Formula,v, the solution forAv2+Bv+C= 0 , where A, B and C are numbers, often called coefficients, is given by :
-B± √B2-4AC
v = ————————
2A In our case,A= 1
B= 4
C=-12 Accordingly,B2-4AC=
16 - (-48) =
64Applying the quadratic formula :
-4 ± √ 64
v=—————
2Can √ 64 be simplified ?
Yes!The prime factorization of 64is
2•2•2•2•2•2
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).
√ 64 =√2•2•2•2•2•2 =2•2•2•√ 1 =
±8 •√ 1 =
±8
So now we are looking at:
v=(-4±8)/2
Two real solutions:
v =(-4+√64)/2=-2+4= 2.000
or:
v =(-4-√64)/2=-2-4= -6.000
Two solutions were found :
- v = 2
- v = -6
