Solution - Quadratic equations
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "24.01" was replaced by "(2401/100)".
Step by step solution :
Step 1 :
2401
Simplify ————
100
Equation at the end of step 1 :
2401
((n2) + 2n) - ———— = 0
100
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using 100 as the denominator :
n2 + 2n (n2 + 2n) • 100
n2 + 2n = ——————— = ———————————————
1 100
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 3 :
Pulling out like terms :
3.1 Pull out like factors :
n2 + 2n = n • (n + 2)
Adding fractions that have a common denominator :
3.2 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:
n • (n+2) • 100 - (2401) 100n2 + 200n - 2401
———————————————————————— = ———————————————————
100 100
Trying to factor by splitting the middle term
3.3 Factoring 100n2 + 200n - 2401
The first term is, 100n2 its coefficient is 100 .
The middle term is, +200n its coefficient is 200 .
The last term, "the constant", is -2401
Step-1 : Multiply the coefficient of the first term by the constant 100 • -2401 = -240100
Step-2 : Find two factors of -240100 whose sum equals the coefficient of the middle term, which is 200 .
-240100 | + | 1 | = | -240099 | ||
-120050 | + | 2 | = | -120048 | ||
-60025 | + | 4 | = | -60021 | ||
-48020 | + | 5 | = | -48015 | ||
-34300 | + | 7 | = | -34293 | ||
-24010 | + | 10 | = | -24000 |
For tidiness, printing of 39 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 3 :
100n2 + 200n - 2401
——————————————————— = 0
100
Step 4 :
When a fraction equals zero :
4.1 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:
100n2+200n-2401
——————————————— • 100 = 0 • 100
100
Now, on the left hand side, the 100 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
100n2+200n-2401 = 0
Parabola, Finding the Vertex :
4.2 Find the Vertex of y = 100n2+200n-2401
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, 100 , 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,An2+Bn+C,the n -coordinate of the vertex is given by -B/(2A) . In our case the n coordinate is -1.0000
Plugging into the parabola formula -1.0000 for n we can calculate the y -coordinate :
y = 100.0 * -1.00 * -1.00 + 200.0 * -1.00 - 2401.0
or y = -2501.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 100n2+200n-2401
Axis of Symmetry (dashed) {n}={-1.00}
Vertex at {n,y} = {-1.00,-2501.00}
n -Intercepts (Roots) :
Root 1 at {n,y} = {-6.00, 0.00}
Root 2 at {n,y} = { 4.00, 0.00}
Solve Quadratic Equation by Completing The Square
4.3 Solving 100n2+200n-2401 = 0 by Completing The Square .
Divide both sides of the equation by 100 to have 1 as the coefficient of the first term :
n2+2n-(2401/100) = 0
Add 2401/100 to both side of the equation :
n2+2n = 2401/100
Now the clever bit: Take the coefficient of n , which is 2 , divide by two, giving 1 , and finally square it giving 1
Add 1 to both sides of the equation :
On the right hand side we have :
2401/100 + 1 or, (2401/100)+(1/1)
The common denominator of the two fractions is 100 Adding (2401/100)+(100/100) gives 2501/100
So adding to both sides we finally get :
n2+2n+1 = 2501/100
Adding 1 has completed the left hand side into a perfect square :
n2+2n+1 =
(n+1) • (n+1) =
(n+1)2
Things which are equal to the same thing are also equal to one another. Since
n2+2n+1 = 2501/100 and
n2+2n+1 = (n+1)2
then, according to the law of transitivity,
(n+1)2 = 2501/100
We'll refer to this Equation as Eq. #4.3.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(n+1)2 is
(n+1)2/2 =
(n+1)1 =
n+1
Now, applying the Square Root Principle to Eq. #4.3.1 we get:
n+1 = √ 2501/100
Subtract 1 from both sides to obtain:
n = -1 + √ 2501/100
Since a square root has two values, one positive and the other negative
n2 + 2n - (2401/100) = 0
has two solutions:
n = -1 + √ 2501/100
or
n = -1 - √ 2501/100
Note that √ 2501/100 can be written as
√ 2501 / √ 100 which is √ 2501 / 10
Solve Quadratic Equation using the Quadratic Formula
4.4 Solving 100n2+200n-2401 = 0 by the Quadratic Formula .
According to the Quadratic Formula, n , the solution for An2+Bn+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
n = ————————
2A
In our case, A = 100
B = 200
C = -2401
Accordingly, B2 - 4AC =
40000 - (-960400) =
1000400
Applying the quadratic formula :
-200 ± √ 1000400
n = ——————————
200
Can √ 1000400 be simplified ?
Yes! The prime factorization of 1000400 is
2•2•2•2•5•5•41•61
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).
√ 1000400 = √ 2•2•2•2•5•5•41•61 =2•2•5•√ 2501 =
± 20 • √ 2501
√ 2501 , rounded to 4 decimal digits, is 50.0100
So now we are looking at:
n = ( -200 ± 20 • 50.010 ) / 200
Two real solutions:
n =(-200+√1000400)/200=-1+1/10√ 2501 = 4.001
or:
n =(-200-√1000400)/200=-1-1/10√ 2501 = -6.001
Two solutions were found :
- n =(-200-√1000400)/200=-1-1/10√ 2501 = -6.001
- n =(-200+√1000400)/200=-1+1/10√ 2501 = 4.001
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