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 :
2/x-2+7/x^2-4-(5/x)=0
Step by step solution :
Step 1 :
5
Simplify —
x
Equation at the end of step 1 :
2 7 5 (((—-2)+————)-4)-— = 0 x (x2) xStep 2 :
7 Simplify —— x2
Equation at the end of step 2 :
2 7 5
(((— - 2) + ——) - 4) - — = 0
x x2 x
Step 3 :
2
Simplify —
x
Equation at the end of step 3 :
2 7 5
(((— - 2) + ——) - 4) - — = 0
x x2 x
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using x as the denominator :
2 2 • x
2 = — = —————
1 x
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 :
4.2 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:
2 - (2 • x) 2 - 2x
——————————— = ——————
x x
Equation at the end of step 4 :
(2 - 2x) 7 5
((———————— + ——) - 4) - — = 0
x x2 x
Step 5 :
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
2 - 2x = -2 • (x - 1)
Calculating the Least Common Multiple :
6.2 Find the Least Common Multiple
The left denominator is : x
The right denominator is : x2
Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
x | 1 | 2 | 2 |
Least Common Multiple:
x2
Calculating Multipliers :
6.3 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 = x
Right_M = L.C.M / R_Deno = 1
Making Equivalent Fractions :
6.4 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 • (x-1) • x —————————————————— = —————————————— L.C.M x2 R. Mult. • R. Num. 7 —————————————————— = —— L.C.M x2
Adding fractions that have a common denominator :
6.5 Adding up the two equivalent fractions
-2 • (x-1) • x + 7 -2x2 + 2x + 7
—————————————————— = —————————————
x2 x2
Equation at the end of step 6 :
(-2x2 + 2x + 7) 5
(——————————————— - 4) - — = 0
x2 x
Step 7 :
Rewriting the whole as an Equivalent Fraction :
7.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using x2 as the denominator :
4 4 • x2
4 = — = ——————
1 x2
Trying to factor by splitting the middle term
7.2 Factoring -2x2 + 2x + 7
The first term is, -2x2 its coefficient is -2 .
The middle term is, +2x its coefficient is 2 .
The last term, "the constant", is +7
Step-1 : Multiply the coefficient of the first term by the constant -2 • 7 = -14
Step-2 : Find two factors of -14 whose sum equals the coefficient of the middle term, which is 2 .
-14 | + | 1 | = | -13 | ||
-7 | + | 2 | = | -5 | ||
-2 | + | 7 | = | 5 | ||
-1 | + | 14 | = | 13 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Adding fractions that have a common denominator :
7.3 Adding up the two equivalent fractions
(-2x2+2x+7) - (4 • x2) -6x2 + 2x + 7
—————————————————————— = —————————————
x2 x2
Equation at the end of step 7 :
(-6x2 + 2x + 7) 5
——————————————— - — = 0
x2 x
Step 8 :
Trying to factor by splitting the middle term
8.1 Factoring -6x2+2x+7
The first term is, -6x2 its coefficient is -6 .
The middle term is, +2x its coefficient is 2 .
The last term, "the constant", is +7
Step-1 : Multiply the coefficient of the first term by the constant -6 • 7 = -42
Step-2 : Find two factors of -42 whose sum equals the coefficient of the middle term, which is 2 .
-42 | + | 1 | = | -41 | ||
-21 | + | 2 | = | -19 | ||
-14 | + | 3 | = | -11 | ||
-7 | + | 6 | = | -1 | ||
-6 | + | 7 | = | 1 | ||
-3 | + | 14 | = | 11 | ||
-2 | + | 21 | = | 19 | ||
-1 | + | 42 | = | 41 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Calculating the Least Common Multiple :
8.2 Find the Least Common Multiple
The left denominator is : x2
The right denominator is : x
Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
x | 2 | 1 | 2 |
Least Common Multiple:
x2
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 = x
Making Equivalent Fractions :
8.4 Rewrite the two fractions into equivalent fractions
L. Mult. • L. Num. (-6x2+2x+7) —————————————————— = ——————————— L.C.M x2 R. Mult. • R. Num. 5 • x —————————————————— = ————— L.C.M x2
Adding fractions that have a common denominator :
8.5 Adding up the two equivalent fractions
(-6x2+2x+7) - (5 • x) -6x2 - 3x + 7
————————————————————— = —————————————
x2 x2
Step 9 :
Pulling out like terms :
9.1 Pull out like factors :
-6x2 - 3x + 7 = -1 • (6x2 + 3x - 7)
Trying to factor by splitting the middle term
9.2 Factoring 6x2 + 3x - 7
The first term is, 6x2 its coefficient is 6 .
The middle term is, +3x its coefficient is 3 .
The last term, "the constant", is -7
Step-1 : Multiply the coefficient of the first term by the constant 6 • -7 = -42
Step-2 : Find two factors of -42 whose sum equals the coefficient of the middle term, which is 3 .
-42 | + | 1 | = | -41 | ||
-21 | + | 2 | = | -19 | ||
-14 | + | 3 | = | -11 | ||
-7 | + | 6 | = | -1 | ||
-6 | + | 7 | = | 1 | ||
-3 | + | 14 | = | 11 | ||
-2 | + | 21 | = | 19 | ||
-1 | + | 42 | = | 41 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 9 :
-6x2 - 3x + 7
————————————— = 0
x2
Step 10 :
When a fraction equals zero :
10.1 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:
-6x2-3x+7
————————— • x2 = 0 • x2
x2
Now, on the left hand side, the x2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
-6x2-3x+7 = 0
Parabola, Finding the Vertex :
10.2 Find the Vertex of y = -6x2-3x+7
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, -6 , 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,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.2500
Plugging into the parabola formula -0.2500 for x we can calculate the y -coordinate :
y = -6.0 * -0.25 * -0.25 - 3.0 * -0.25 + 7.0
or y = 7.375
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -6x2-3x+7
Axis of Symmetry (dashed) {x}={-0.25}
Vertex at {x,y} = {-0.25, 7.38}
x -Intercepts (Roots) :
Root 1 at {x,y} = { 0.86, 0.00}
Root 2 at {x,y} = {-1.36, 0.00}
Solve Quadratic Equation by Completing The Square
10.3 Solving -6x2-3x+7 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
6x2+3x-7 = 0 Divide both sides of the equation by 6 to have 1 as the coefficient of the first term :
x2+(1/2)x-(7/6) = 0
Add 7/6 to both side of the equation :
x2+(1/2)x = 7/6
Now the clever bit: Take the coefficient of x , which is 1/2 , divide by two, giving 1/4 , and finally square it giving 1/16
Add 1/16 to both sides of the equation :
On the right hand side we have :
7/6 + 1/16 The common denominator of the two fractions is 48 Adding (56/48)+(3/48) gives 59/48
So adding to both sides we finally get :
x2+(1/2)x+(1/16) = 59/48
Adding 1/16 has completed the left hand side into a perfect square :
x2+(1/2)x+(1/16) =
(x+(1/4)) • (x+(1/4)) =
(x+(1/4))2
Things which are equal to the same thing are also equal to one another. Since
x2+(1/2)x+(1/16) = 59/48 and
x2+(1/2)x+(1/16) = (x+(1/4))2
then, according to the law of transitivity,
(x+(1/4))2 = 59/48
We'll refer to this Equation as Eq. #10.3.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x+(1/4))2 is
(x+(1/4))2/2 =
(x+(1/4))1 =
x+(1/4)
Now, applying the Square Root Principle to Eq. #10.3.1 we get:
x+(1/4) = √ 59/48
Subtract 1/4 from both sides to obtain:
x = -1/4 + √ 59/48
Since a square root has two values, one positive and the other negative
x2 + (1/2)x - (7/6) = 0
has two solutions:
x = -1/4 + √ 59/48
or
x = -1/4 - √ 59/48
Note that √ 59/48 can be written as
√ 59 / √ 48
Solve Quadratic Equation using the Quadratic Formula
10.4 Solving -6x2-3x+7 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = -6
B = -3
C = 7
Accordingly, B2 - 4AC =
9 - (-168) =
177
Applying the quadratic formula :
3 ± √ 177
x = —————
-12
√ 177 , rounded to 4 decimal digits, is 13.3041
So now we are looking at:
x = ( 3 ± 13.304 ) / -12
Two real solutions:
x =(3+√177)/-12=1/-4-1/12√ 177 = -1.359
or:
x =(3-√177)/-12=1/-4+1/12√ 177 = 0.859
Two solutions were found :
- x =(3-√177)/-12=1/-4+1/12√ 177 = 0.859
- x =(3+√177)/-12=1/-4-1/12√ 177 = -1.359
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