Solution - Adding, subtracting and finding the least common multiple
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "30.5" was replaced by "(305/10)". 2 more similar replacement(s)
Step by step solution :
Step 1 :
61
Simplify ——
2
Equation at the end of step 1 :
49 61 ((——•(x2))-(——•x))-183 = 0 10 2Step 2 :
49 Simplify —— 10
Equation at the end of step 2 :
49 61x
((—— • x2) - ———) - 183 = 0
10 2
Step 3 :
Equation at the end of step 3 :
49x2 61x
(———— - ———) - 183 = 0
10 2
Step 4 :
Calculating the Least Common Multiple :
4.1 Find the Least Common Multiple
The left denominator is : 10
The right denominator is : 2
Prime Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
2 | 1 | 1 | 1 |
5 | 1 | 0 | 1 |
Product of all Prime Factors | 10 | 2 | 10 |
Least Common Multiple:
10
Calculating Multipliers :
4.2 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 = 5
Making Equivalent Fractions :
4.3 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. 49x2 —————————————————— = ———— L.C.M 10 R. Mult. • R. Num. 61x • 5 —————————————————— = ——————— L.C.M 10
Adding fractions that have a common denominator :
4.4 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:
49x2 - (61x • 5) 49x2 - 305x
———————————————— = ———————————
10 10
Equation at the end of step 4 :
(49x2 - 305x)
————————————— - 183 = 0
10
Step 5 :
Rewriting the whole as an Equivalent Fraction :
5.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 10 as the denominator :
183 183 • 10
183 = ——— = ————————
1 10
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
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
49x2 - 305x = x • (49x - 305)
Adding fractions that have a common denominator :
6.2 Adding up the two equivalent fractions
x • (49x-305) - (183 • 10) 49x2 - 305x - 1830
—————————————————————————— = ——————————————————
10 10
Trying to factor by splitting the middle term
6.3 Factoring 49x2 - 305x - 1830
The first term is, 49x2 its coefficient is 49 .
The middle term is, -305x its coefficient is -305 .
The last term, "the constant", is -1830
Step-1 : Multiply the coefficient of the first term by the constant 49 • -1830 = -89670
Step-2 : Find two factors of -89670 whose sum equals the coefficient of the middle term, which is -305 .
-89670 | + | 1 | = | -89669 | ||
-44835 | + | 2 | = | -44833 | ||
-29890 | + | 3 | = | -29887 | ||
-17934 | + | 5 | = | -17929 | ||
-14945 | + | 6 | = | -14939 | ||
-12810 | + | 7 | = | -12803 |
For tidiness, printing of 42 lines which failed to find two such factors, was suppressed
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 6 :
49x2 - 305x - 1830
—————————————————— = 0
10
Step 7 :
When a fraction equals zero :
7.1 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:
49x2-305x-1830
—————————————— • 10 = 0 • 10
10
Now, on the left hand side, the 10 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
49x2-305x-1830 = 0
Parabola, Finding the Vertex :
7.2 Find the Vertex of y = 49x2-305x-1830
Parabolas 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, 49 , 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,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 3.1122
Plugging into the parabola formula 3.1122 for x we can calculate the y -coordinate :
y = 49.0 * 3.11 * 3.11 - 305.0 * 3.11 - 1830.0
or y = -2304.617
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 49x2-305x-1830
Axis of Symmetry (dashed) {x}={ 3.11}
Vertex at {x,y} = { 3.11,-2304.62}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-3.75, 0.00}
Root 2 at {x,y} = { 9.97, 0.00}
Solve Quadratic Equation by Completing The Square
7.3 Solving 49x2-305x-1830 = 0 by Completing The Square .
Divide both sides of the equation by 49 to have 1 as the coefficient of the first term :
x2-(305/49)x-(1830/49) = 0
Add 1830/49 to both side of the equation :
x2-(305/49)x = 1830/49
Now the clever bit: Take the coefficient of x , which is 305/49 , divide by two, giving 305/98 , and finally square it giving 93025/9604
Add 93025/9604 to both sides of the equation :
On the right hand side we have :
1830/49 + 93025/9604 The common denominator of the two fractions is 9604 Adding (358680/9604)+(93025/9604) gives 451705/9604
So adding to both sides we finally get :
x2-(305/49)x+(93025/9604) = 451705/9604
Adding 93025/9604 has completed the left hand side into a perfect square :
x2-(305/49)x+(93025/9604) =
(x-(305/98)) • (x-(305/98)) =
(x-(305/98))2
Things which are equal to the same thing are also equal to one another. Since
x2-(305/49)x+(93025/9604) = 451705/9604 and
x2-(305/49)x+(93025/9604) = (x-(305/98))2
then, according to the law of transitivity,
(x-(305/98))2 = 451705/9604
We'll refer to this Equation as Eq. #7.3.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(305/98))2 is
(x-(305/98))2/2 =
(x-(305/98))1 =
x-(305/98)
Now, applying the Square Root Principle to Eq. #7.3.1 we get:
x-(305/98) = √ 451705/9604
Add 305/98 to both sides to obtain:
x = 305/98 + √ 451705/9604
Since a square root has two values, one positive and the other negative
x2 - (305/49)x - (1830/49) = 0
has two solutions:
x = 305/98 + √ 451705/9604
or
x = 305/98 - √ 451705/9604
Note that √ 451705/9604 can be written as
√ 451705 / √ 9604 which is √ 451705 / 98
Solve Quadratic Equation using the Quadratic Formula
7.4 Solving 49x2-305x-1830 = 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 = 49
B = -305
C = -1830
Accordingly, B2 - 4AC =
93025 - (-358680) =
451705
Applying the quadratic formula :
305 ± √ 451705
x = ————————
98
√ 451705 , rounded to 4 decimal digits, is 672.0900
So now we are looking at:
x = ( 305 ± 672.090 ) / 98
Two real solutions:
x =(305+√451705)/98= 9.970
or:
x =(305-√451705)/98=-3.746
Two solutions were found :
- x =(305-√451705)/98=-3.746
- x =(305+√451705)/98= 9.970
How did we do?
Please leave us feedback.