Solution - Reducing fractions to their lowest terms
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "3022.1" was replaced by "(30221/10)".
Step by step solution :
Step 1 :
30221
Simplify —————
10
Equation at the end of step 1 :
30221 ((9996 • (x2)) - 11029x) + ————— = 0 10Step 2 :
Equation at the end of step 2 :
30221
((22•3•72•17x2) - 11029x) + ————— = 0
10
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Adding a fraction to a whole
Rewrite the whole as a fraction using 10 as the denominator :
9996x2 - 11029x (9996x2 - 11029x) • 10
9996x2 - 11029x = ——————————————— = ——————————————————————
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 4 :
Pulling out like terms :
4.1 Pull out like factors :
9996x2 - 11029x = x • (9996x - 11029)
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:
x • (9996x-11029) • 10 + 30221 99960x2 - 110290x + 30221
—————————————————————————————— = —————————————————————————
10 10
Trying to factor by splitting the middle term
4.3 Factoring 99960x2 - 110290x + 30221
The first term is, 99960x2 its coefficient is 99960 .
The middle term is, -110290x its coefficient is -110290 .
The last term, "the constant", is +30221
Step-1 : Multiply the coefficient of the first term by the constant
Numbers too big. Method shall not be applied
Equation at the end of step 4 :
99960x2 - 110290x + 30221
————————————————————————— = 0
10
Step 5 :
When a fraction equals zero :
5.1 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:
99960x2-110290x+30221
————————————————————— • 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 :
99960x2-110290x+30221 = 0
Parabola, Finding the Vertex :
5.2 Find the Vertex of y = 99960x2-110290x+30221
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, 99960 , 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 0.5517
Plugging into the parabola formula 0.5517 for x we can calculate the y -coordinate :
y = 99960.0 * 0.55 * 0.55 - 110290.0 * 0.55 + 30221.0
or y = -200.879
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 99960x2-110290x+30221
Axis of Symmetry (dashed) {x}={ 0.55}
Vertex at {x,y} = { 0.55,-200.88}
x -Intercepts (Roots) :
Root 1 at {x,y} = { 0.51, 0.00}
Root 2 at {x,y} = { 0.60, 0.00}
Solve Quadratic Equation using the Quadratic Formula
5.3 Solving 99960x2-110290x+30221 = 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 = 99960.00
B = -110290.00
C = 30221.00
B2 = 12163884100.00
4AC = 12083564640.00
B2 - 4AC = 80319460.00
SQRT(B2-4AC) = 8962.11
x=( 110290.00 ± 8962.11) / 199920.00
x = 0.59650
x = 0.50684
Two solutions were found :
- x = 0.50684
- x = 0.59650
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