Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-200x-2400
The first term is, x2 its coefficient is 1 .
The middle term is, -200x its coefficient is -200 .
The last term, "the constant", is -2400
Step-1 : Multiply the coefficient of the first term by the constant 1 • -2400 = -2400
Step-2 : Find two factors of -2400 whose sum equals the coefficient of the middle term, which is -200 .
| -2400 | + | 1 | = | -2399 | ||
| -1200 | + | 2 | = | -1198 | ||
| -800 | + | 3 | = | -797 | ||
| -600 | + | 4 | = | -596 | ||
| -480 | + | 5 | = | -475 | ||
| -400 | + | 6 | = | -394 |
For tidiness, printing of 30 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 - 200x - 2400 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2-200x-2400
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, 1 , 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 100.0000
Plugging into the parabola formula 100.0000 for x we can calculate the y -coordinate :
y = 1.0 * 100.00 * 100.00 - 200.0 * 100.00 - 2400.0
or y = -12400.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-200x-2400
Axis of Symmetry (dashed) {x}={100.00}
Vertex at {x,y} = {100.00,-12400.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-11.36, 0.00}
Root 2 at {x,y} = {211.36, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving x2-200x-2400 = 0 by Completing The Square .
Add 2400 to both side of the equation :
x2-200x = 2400
Now the clever bit: Take the coefficient of x , which is 200 , divide by two, giving 100 , and finally square it giving 10000
Add 10000 to both sides of the equation :
On the right hand side we have :
2400 + 10000 or, (2400/1)+(10000/1)
The common denominator of the two fractions is 1 Adding (2400/1)+(10000/1) gives 12400/1
So adding to both sides we finally get :
x2-200x+10000 = 12400
Adding 10000 has completed the left hand side into a perfect square :
x2-200x+10000 =
(x-100) • (x-100) =
(x-100)2
Things which are equal to the same thing are also equal to one another. Since
x2-200x+10000 = 12400 and
x2-200x+10000 = (x-100)2
then, according to the law of transitivity,
(x-100)2 = 12400
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-100)2 is
(x-100)2/2 =
(x-100)1 =
x-100
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-100 = √ 12400
Add 100 to both sides to obtain:
x = 100 + √ 12400
Since a square root has two values, one positive and the other negative
x2 - 200x - 2400 = 0
has two solutions:
x = 100 + √ 12400
or
x = 100 - √ 12400
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving x2-200x-2400 = 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 = -200
C = -2400
Accordingly, B2 - 4AC =
40000 - (-9600) =
49600
Applying the quadratic formula :
200 ± √ 49600
x = ————————
2
Can √ 49600 be simplified ?
Yes! The prime factorization of 49600 is
2•2•2•2•2•2•5•5•31
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).
√ 49600 = √ 2•2•2•2•2•2•5•5•31 =2•2•2•5•√ 31 =
± 40 • √ 31
√ 31 , rounded to 4 decimal digits, is 5.5678
So now we are looking at:
x = ( 200 ± 40 • 5.568 ) / 2
Two real solutions:
x =(200+√49600)/2=100+20√ 31 = 211.355
or:
x =(200-√49600)/2=100-20√ 31 = -11.355
Two solutions were found :
- x =(200-√49600)/2=100-20√ 31 = -11.355
- x =(200+√49600)/2=100+20√ 31 = 211.355
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