Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring -x2+120x+5000
The first term is, -x2 its coefficient is -1 .
The middle term is, +120x its coefficient is 120 .
The last term, "the constant", is +5000
Step-1 : Multiply the coefficient of the first term by the constant -1 • 5000 = -5000
Step-2 : Find two factors of -5000 whose sum equals the coefficient of the middle term, which is 120 .
| -5000 | + | 1 | = | -4999 | ||
| -2500 | + | 2 | = | -2498 | ||
| -1250 | + | 4 | = | -1246 | ||
| -1000 | + | 5 | = | -995 | ||
| -625 | + | 8 | = | -617 | ||
| -500 | + | 10 | = | -490 |
For tidiness, printing of 14 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 1 :
-x2 + 120x + 5000 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = -x2+120x+5000
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, -1 , 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 60.0000
Plugging into the parabola formula 60.0000 for x we can calculate the y -coordinate :
y = -1.0 * 60.00 * 60.00 + 120.0 * 60.00 + 5000.0
or y = 8600.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -x2+120x+5000
Axis of Symmetry (dashed) {x}={60.00}
Vertex at {x,y} = {60.00,8600.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {152.74, 0.00}
Root 2 at {x,y} = {-32.74, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving -x2+120x+5000 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
x2-120x-5000 = 0 Add 5000 to both side of the equation :
x2-120x = 5000
Now the clever bit: Take the coefficient of x , which is 120 , divide by two, giving 60 , and finally square it giving 3600
Add 3600 to both sides of the equation :
On the right hand side we have :
5000 + 3600 or, (5000/1)+(3600/1)
The common denominator of the two fractions is 1 Adding (5000/1)+(3600/1) gives 8600/1
So adding to both sides we finally get :
x2-120x+3600 = 8600
Adding 3600 has completed the left hand side into a perfect square :
x2-120x+3600 =
(x-60) • (x-60) =
(x-60)2
Things which are equal to the same thing are also equal to one another. Since
x2-120x+3600 = 8600 and
x2-120x+3600 = (x-60)2
then, according to the law of transitivity,
(x-60)2 = 8600
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-60)2 is
(x-60)2/2 =
(x-60)1 =
x-60
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-60 = √ 8600
Add 60 to both sides to obtain:
x = 60 + √ 8600
Since a square root has two values, one positive and the other negative
x2 - 120x - 5000 = 0
has two solutions:
x = 60 + √ 8600
or
x = 60 - √ 8600
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving -x2+120x+5000 = 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 = -1
B = 120
C = 5000
Accordingly, B2 - 4AC =
14400 - (-20000) =
34400
Applying the quadratic formula :
-120 ± √ 34400
x = ————————
-2
Can √ 34400 be simplified ?
Yes! The prime factorization of 34400 is
2•2•2•2•2•5•5•43
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).
√ 34400 = √ 2•2•2•2•2•5•5•43 =2•2•5•√ 86 =
± 20 • √ 86
√ 86 , rounded to 4 decimal digits, is 9.2736
So now we are looking at:
x = ( -120 ± 20 • 9.274 ) / -2
Two real solutions:
x =(-120+√34400)/-2=60-10√ 86 = -32.736
or:
x =(-120-√34400)/-2=60+10√ 86 = 152.736
Two solutions were found :
- x =(-120-√34400)/-2=60+10√ 86 = 152.736
- x =(-120+√34400)/-2=60-10√ 86 = -32.736
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