Solution - Adding, subtracting and finding the least common multiple
Step by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
4/a^2-1+a-2/a-1-(a-3/a+1)=0
Step by step solution :
Step 1 :
3
Simplify —
a
Equation at the end of step 1 :
4 2 3
((((————-1)+a)-—)-1)-((a-—)+1) = 0
(a2) a a
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using a as the denominator :
a a • a
a = — = —————
1 a
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 :
2.2 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:
a • a - (3) a2 - 3
——————————— = ——————
a a
Equation at the end of step 2 :
4 2 (a2-3)
((((————-1)+a)-—)-1)-(——————+1) = 0
(a2) a a
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Adding a whole to a fraction
Rewrite the whole as a fraction using a as the denominator :
1 1 • a
1 = — = —————
1 a
Trying to factor as a Difference of Squares :
3.2 Factoring: a2 - 3
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 3 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Adding fractions that have a common denominator :
3.3 Adding up the two equivalent fractions
(a2-3) + a a2 + a - 3
—————————— = ——————————
a a
Equation at the end of step 3 :
4 2 (a2+a-3)
((((————-1)+a)-—)-1)-———————— = 0
(a2) a a
Step 4 :
2
Simplify —
a
Equation at the end of step 4 :
4 2 (a2+a-3) ((((————-1)+a)-—)-1)-———————— = 0 (a2) a aStep 5 :
4 Simplify —— a2
Equation at the end of step 5 :
4 2 (a2+a-3)
((((——-1)+a)-—)-1)-———————— = 0
a2 a a
Step 6 :
Rewriting the whole as an Equivalent Fraction :
6.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using a2 as the denominator :
1 1 • a2
1 = — = ——————
1 a2
Adding fractions that have a common denominator :
6.2 Adding up the two equivalent fractions
4 - (a2) 4 - a2
———————— = ——————
a2 a2
Equation at the end of step 6 :
(4 - a2) 2 (a2 + a - 3)
(((———————— + a) - —) - 1) - ———————————— = 0
a2 a a
Step 7 :
Rewriting the whole as an Equivalent Fraction :
7.1 Adding a whole to a fraction
Rewrite the whole as a fraction using a2 as the denominator :
a a • a2
a = — = ——————
1 a2
Trying to factor as a Difference of Squares :
7.2 Factoring: 4 - a2
Check : 4 is the square of 2
Check : a2 is the square of a1
Factorization is : (2 + a) • (2 - a)
Adding fractions that have a common denominator :
7.3 Adding up the two equivalent fractions
(a+2) • (2-a) + a • a2 a3 - a2 + 4
—————————————————————— = ———————————
a2 a2
Equation at the end of step 7 :
(a3 - a2 + 4) 2 (a2 + a - 3)
((————————————— - —) - 1) - ———————————— = 0
a2 a a
Step 8 :
Polynomial Roots Calculator :
8.1 Find roots (zeroes) of : F(a) = a3-a2+4
Polynomial Roots Calculator is a set of methods aimed at finding values of a for which F(a)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers a which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is 4.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 2.00 | ||||||
| -2 | 1 | -2.00 | -8.00 | ||||||
| -4 | 1 | -4.00 | -76.00 | ||||||
| 1 | 1 | 1.00 | 4.00 | ||||||
| 2 | 1 | 2.00 | 8.00 | ||||||
| 4 | 1 | 4.00 | 52.00 |
Polynomial Roots Calculator found no rational roots
Calculating the Least Common Multiple :
8.2 Find the Least Common Multiple
The left denominator is : a2
The right denominator is : a
| Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| a | 2 | 1 | 2 |
Least Common Multiple:
a2
Calculating Multipliers :
8.3 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 = 1
Right_M = L.C.M / R_Deno = a
Making Equivalent Fractions :
8.4 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. (a3-a2+4) —————————————————— = ————————— L.C.M a2 R. Mult. • R. Num. 2 • a —————————————————— = ————— L.C.M a2
Adding fractions that have a common denominator :
8.5 Adding up the two equivalent fractions
(a3-a2+4) - (2 • a) a3 - a2 - 2a + 4
——————————————————— = ————————————————
a2 a2
Equation at the end of step 8 :
(a3 - a2 - 2a + 4) (a2 + a - 3)
(—————————————————— - 1) - ———————————— = 0
a2 a
Step 9 :
Rewriting the whole as an Equivalent Fraction :
9.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using a2 as the denominator :
1 1 • a2
1 = — = ——————
1 a2
Checking for a perfect cube :
9.2 a3 - a2 - 2a + 4 is not a perfect cube
Trying to factor by pulling out :
9.3 Factoring: a3 - a2 - 2a + 4
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -2a + 4
Group 2: a3 - a2
Pull out from each group separately :
Group 1: (a - 2) • (-2)
Group 2: (a - 1) • (a2)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
9.4 Find roots (zeroes) of : F(a) = a3 - a2 - 2a + 4
See theory in step 8.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 4.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 4.00 | ||||||
| -2 | 1 | -2.00 | -4.00 | ||||||
| -4 | 1 | -4.00 | -68.00 | ||||||
| 1 | 1 | 1.00 | 2.00 | ||||||
| 2 | 1 | 2.00 | 4.00 | ||||||
| 4 | 1 | 4.00 | 44.00 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
9.5 Adding up the two equivalent fractions
(a3-a2-2a+4) - (a2) a3 - 2a2 - 2a + 4
——————————————————— = —————————————————
a2 a2
Equation at the end of step 9 :
(a3 - 2a2 - 2a + 4) (a2 + a - 3)
——————————————————— - ———————————— = 0
a2 a
Step 10 :
Checking for a perfect cube :
10.1 a3-2a2-2a+4 is not a perfect cube
Trying to factor by pulling out :
10.2 Factoring: a3-2a2-2a+4
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -2a+4
Group 2: -2a2+a3
Pull out from each group separately :
Group 1: (a-2) • (-2)
Group 2: (a-2) • (a2)
-------------------
Add up the two groups :
(a-2) • (a2-2)
Which is the desired factorization
Trying to factor as a Difference of Squares :
10.3 Factoring: a2-2
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Trying to factor by splitting the middle term
10.4 Factoring a2+a-3
The first term is, a2 its coefficient is 1 .
The middle term is, +a its coefficient is 1 .
The last term, "the constant", is -3
Step-1 : Multiply the coefficient of the first term by the constant 1 • -3 = -3
Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is 1 .
| -3 | + | 1 | = | -2 | ||
| -1 | + | 3 | = | 2 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Calculating the Least Common Multiple :
10.5 Find the Least Common Multiple
The left denominator is : a2
The right denominator is : a
| Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| a | 2 | 1 | 2 |
Least Common Multiple:
a2
Calculating Multipliers :
10.6 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 = 1
Right_M = L.C.M / R_Deno = a
Making Equivalent Fractions :
10.7 Rewrite the two fractions into equivalent fractions
L. Mult. • L. Num. (a2-2) • (a-2) —————————————————— = —————————————— L.C.M a2 R. Mult. • R. Num. (a2+a-3) • a —————————————————— = ———————————— L.C.M a2
Adding fractions that have a common denominator :
10.8 Adding up the two equivalent fractions
(a2-2) • (a-2) - ((a2+a-3) • a) -3a2 + a + 4
——————————————————————————————— = ————————————
a2 a2
Trying to factor by splitting the middle term
10.9 Factoring -3a2 + a + 4
The first term is, -3a2 its coefficient is -3 .
The middle term is, +a its coefficient is 1 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant -3 • 4 = -12
Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is 1 .
| -12 | + | 1 | = | -11 | ||
| -6 | + | 2 | = | -4 | ||
| -4 | + | 3 | = | -1 | ||
| -3 | + | 4 | = | 1 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 4
-3a2 - 3a + 4a + 4
Step-4 : Add up the first 2 terms, pulling out like factors :
-3a • (a+1)
Add up the last 2 terms, pulling out common factors :
4 • (a+1)
Step-5 : Add up the four terms of step 4 :
(-3a+4) • (a+1)
Which is the desired factorization
Equation at the end of step 10 :
(a + 1) • (4 - 3a)
—————————————————— = 0
a2
Step 11 :
When a fraction equals zero :
11.1 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:
(a+1)•(4-3a)
———————————— • a2 = 0 • a2
a2
Now, on the left hand side, the a2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
(a+1) • (4-3a) = 0
Theory - Roots of a product :
11.2 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 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
11.3 Solve : a+1 = 0
Subtract 1 from both sides of the equation :
a = -1
Solving a Single Variable Equation :
11.4 Solve : -3a+4 = 0
Subtract 4 from both sides of the equation :
-3a = -4
Multiply both sides of the equation by (-1) : 3a = 4
Divide both sides of the equation by 3:
a = 4/3 = 1.333
Supplement : Solving Quadratic Equation Directly
Solving -3a2+a+4 = 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 :
12.1 Find the Vertex of y = -3a2+a+4
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens down and accordingly has a highest point (AKA absolute maximum) . We know this even before plotting "y" because the coefficient of the first term, -3 , is negative (smaller 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,Aa2+Ba+C,the a -coordinate of the vertex is given by -B/(2A) . In our case the a coordinate is 0.1667
Plugging into the parabola formula 0.1667 for a we can calculate the y -coordinate :
y = -3.0 * 0.17 * 0.17 + 1.0 * 0.17 + 4.0
or y = 4.083
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -3a2+a+4
Axis of Symmetry (dashed) {a}={ 0.17}
Vertex at {a,y} = { 0.17, 4.08}
a -Intercepts (Roots) :
Root 1 at {a,y} = { 1.33, 0.00}
Root 2 at {a,y} = {-1.00, 0.00}
Solve Quadratic Equation by Completing The Square
12.2 Solving -3a2+a+4 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
3a2-a-4 = 0 Divide both sides of the equation by 3 to have 1 as the coefficient of the first term :
a2-(1/3)a-(4/3) = 0
Add 4/3 to both side of the equation :
a2-(1/3)a = 4/3
Now the clever bit: Take the coefficient of a , which is 1/3 , divide by two, giving 1/6 , and finally square it giving 1/36
Add 1/36 to both sides of the equation :
On the right hand side we have :
4/3 + 1/36 The common denominator of the two fractions is 36 Adding (48/36)+(1/36) gives 49/36
So adding to both sides we finally get :
a2-(1/3)a+(1/36) = 49/36
Adding 1/36 has completed the left hand side into a perfect square :
a2-(1/3)a+(1/36) =
(a-(1/6)) • (a-(1/6)) =
(a-(1/6))2
Things which are equal to the same thing are also equal to one another. Since
a2-(1/3)a+(1/36) = 49/36 and
a2-(1/3)a+(1/36) = (a-(1/6))2
then, according to the law of transitivity,
(a-(1/6))2 = 49/36
We'll refer to this Equation as Eq. #12.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(a-(1/6))2 is
(a-(1/6))2/2 =
(a-(1/6))1 =
a-(1/6)
Now, applying the Square Root Principle to Eq. #12.2.1 we get:
a-(1/6) = √ 49/36
Add 1/6 to both sides to obtain:
a = 1/6 + √ 49/36
Since a square root has two values, one positive and the other negative
a2 - (1/3)a - (4/3) = 0
has two solutions:
a = 1/6 + √ 49/36
or
a = 1/6 - √ 49/36
Note that √ 49/36 can be written as
√ 49 / √ 36 which is 7 / 6
Solve Quadratic Equation using the Quadratic Formula
12.3 Solving -3a2+a+4 = 0 by the Quadratic Formula .
According to the Quadratic Formula, a , the solution for Aa2+Ba+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
a = ————————
2A
In our case, A = -3
B = 1
C = 4
Accordingly, B2 - 4AC =
1 - (-48) =
49
Applying the quadratic formula :
-1 ± √ 49
a = —————
-6
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:
a = ( -1 ± 7) / -6
Two real solutions:
a =(-1+√49)/-6=(1-7)/6= -1.000
or:
a =(-1-√49)/-6=(1+7)/6= 1.333
Two solutions were found :
- a = 4/3 = 1.333
- a = -1
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