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): "117.6" was replaced by "(1176/10)". 2 more similar replacement(s)
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
160*x-(49/10)*x^2-((1176/10))=0
Step by step solution :
Step 1 :
588
Simplify ———
5
Equation at the end of step 1 :
49 588 (160x - (—— • (x2))) - ——— = 0 10 5Step 2 :
49 Simplify —— 10
Equation at the end of step 2 :
49 588
(160x - (—— • x2)) - ——— = 0
10 5
Step 3 :
Equation at the end of step 3 :
49x2 588
(160x - ————) - ——— = 0
10 5
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using 10 as the denominator :
160x 160x • 10
160x = ———— = —————————
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
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:
160x • 10 - (49x2) 1600x - 49x2
—————————————————— = ————————————
10 10
Equation at the end of step 4 :
(1600x - 49x2) 588
—————————————— - ——— = 0
10 5
Step 5 :
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
1600x - 49x2 = -x • (49x - 1600)
Calculating the Least Common Multiple :
6.2 Find the Least Common Multiple
The left denominator is : 10
The right denominator is : 5
Prime Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
2 | 1 | 0 | 1 |
5 | 1 | 1 | 1 |
Product of all Prime Factors | 10 | 5 | 10 |
Least Common Multiple:
10
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 = 1
Right_M = L.C.M / R_Deno = 2
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. -x • (49x-1600) —————————————————— = ——————————————— L.C.M 10 R. Mult. • R. Num. 588 • 2 —————————————————— = ——————— L.C.M 10
Adding fractions that have a common denominator :
6.5 Adding up the two equivalent fractions
-x • (49x-1600) - (588 • 2) -49x2 + 1600x - 1176
——————————————————————————— = ————————————————————
10 10
Step 7 :
Pulling out like terms :
7.1 Pull out like factors :
-49x2 + 1600x - 1176 = -1 • (49x2 - 1600x + 1176)
Trying to factor by splitting the middle term
7.2 Factoring 49x2 - 1600x + 1176
The first term is, 49x2 its coefficient is 49 .
The middle term is, -1600x its coefficient is -1600 .
The last term, "the constant", is +1176
Step-1 : Multiply the coefficient of the first term by the constant 49 • 1176 = 57624
Step-2 : Find two factors of 57624 whose sum equals the coefficient of the middle term, which is -1600 .
-57624 | + | -1 | = | -57625 | ||
-28812 | + | -2 | = | -28814 | ||
-19208 | + | -3 | = | -19211 | ||
-14406 | + | -4 | = | -14410 | ||
-9604 | + | -6 | = | -9610 | ||
-8232 | + | -7 | = | -8239 |
For tidiness, printing of 74 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 7 :
-49x2 + 1600x - 1176
———————————————————— = 0
10
Step 8 :
When a fraction equals zero :
8.1 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+1600x-1176
———————————————— • 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+1600x-1176 = 0
Parabola, Finding the Vertex :
8.2 Find the Vertex of y = -49x2+1600x-1176
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, -49 , 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 16.3265
Plugging into the parabola formula 16.3265 for x we can calculate the y -coordinate :
y = -49.0 * 16.33 * 16.33 + 1600.0 * 16.33 - 1176.0
or y = 11885.224
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -49x2+1600x-1176
Axis of Symmetry (dashed) {x}={16.33}
Vertex at {x,y} = {16.33,11885.22}
x -Intercepts (Roots) :
Root 1 at {x,y} = {31.90, 0.00}
Root 2 at {x,y} = { 0.75, 0.00}
Solve Quadratic Equation by Completing The Square
8.3 Solving -49x2+1600x-1176 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
49x2-1600x+1176 = 0 Divide both sides of the equation by 49 to have 1 as the coefficient of the first term :
x2-(1600/49)x+24 = 0
Subtract 24 from both side of the equation :
x2-(1600/49)x = -24
Now the clever bit: Take the coefficient of x , which is 1600/49 , divide by two, giving 800/49 , and finally square it giving 800/49
Add 800/49 to both sides of the equation :
On the right hand side we have :
-24 + 800/49 or, (-24/1)+(800/49)
The common denominator of the two fractions is 49 Adding (-1176/49)+(800/49) gives -376/49
So adding to both sides we finally get :
x2-(1600/49)x+(800/49) = -376/49
Adding 800/49 has completed the left hand side into a perfect square :
x2-(1600/49)x+(800/49) =
(x-(800/49)) • (x-(800/49)) =
(x-(800/49))2
Things which are equal to the same thing are also equal to one another. Since
x2-(1600/49)x+(800/49) = -376/49 and
x2-(1600/49)x+(800/49) = (x-(800/49))2
then, according to the law of transitivity,
(x-(800/49))2 = -376/49
We'll refer to this Equation as Eq. #8.3.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(800/49))2 is
(x-(800/49))2/2 =
(x-(800/49))1 =
x-(800/49)
Now, applying the Square Root Principle to Eq. #8.3.1 we get:
x-(800/49) = √ -376/49
Add 800/49 to both sides to obtain:
x = 800/49 + √ -376/49
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Since a square root has two values, one positive and the other negative
x2 - (1600/49)x + 24 = 0
has two solutions:
x = 800/49 + √ 376/49 • i
or
x = 800/49 - √ 376/49 • i
Note that √ 376/49 can be written as
√ 376 / √ 49 which is √ 376 / 7
Solve Quadratic Equation using the Quadratic Formula
8.4 Solving -49x2+1600x-1176 = 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 = 1600
C = -1176
Accordingly, B2 - 4AC =
2560000 - 230496 =
2329504
Applying the quadratic formula :
-1600 ± √ 2329504
x = ——————————
-98
Can √ 2329504 be simplified ?
Yes! The prime factorization of 2329504 is
2•2•2•2•2•72797
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).
√ 2329504 = √ 2•2•2•2•2•72797 =2•2•√ 145594 =
± 4 • √ 145594
√ 145594 , rounded to 4 decimal digits, is 381.5678
So now we are looking at:
x = ( -1600 ± 4 • 381.568 ) / -98
Two real solutions:
x =(-1600+√2329504)/-98=(800-2√ 145594 )/49= 0.752
or:
x =(-1600-√2329504)/-98=(800+2√ 145594 )/49= 31.901
Two solutions were found :
- x =(-1600-√2329504)/-98=(800+2√ 145594 )/49= 31.901
- x =(-1600+√2329504)/-98=(800-2√ 145594 )/49= 0.752
How did we do?
Please leave us feedback.