Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-300x+10000
The first term is, x2 its coefficient is 1 .
The middle term is, -300x its coefficient is -300 .
The last term, "the constant", is +10000
Step-1 : Multiply the coefficient of the first term by the constant 1 • 10000 = 10000
Step-2 : Find two factors of 10000 whose sum equals the coefficient of the middle term, which is -300 .
| -10000 | + | -1 | = | -10001 | ||
| -5000 | + | -2 | = | -5002 | ||
| -2500 | + | -4 | = | -2504 | ||
| -2000 | + | -5 | = | -2005 | ||
| -1250 | + | -8 | = | -1258 | ||
| -1000 | + | -10 | = | -1010 |
For tidiness, printing of 44 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 - 300x + 10000 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2-300x+10000
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 150.0000
Plugging into the parabola formula 150.0000 for x we can calculate the y -coordinate :
y = 1.0 * 150.00 * 150.00 - 300.0 * 150.00 + 10000.0
or y = -12500.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-300x+10000
Axis of Symmetry (dashed) {x}={150.00}
Vertex at {x,y} = {150.00,-12500.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {38.20, 0.00}
Root 2 at {x,y} = {261.80, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving x2-300x+10000 = 0 by Completing The Square .
Subtract 10000 from both side of the equation :
x2-300x = -10000
Now the clever bit: Take the coefficient of x , which is 300 , divide by two, giving 150 , and finally square it giving 22500
Add 22500 to both sides of the equation :
On the right hand side we have :
-10000 + 22500 or, (-10000/1)+(22500/1)
The common denominator of the two fractions is 1 Adding (-10000/1)+(22500/1) gives 12500/1
So adding to both sides we finally get :
x2-300x+22500 = 12500
Adding 22500 has completed the left hand side into a perfect square :
x2-300x+22500 =
(x-150) • (x-150) =
(x-150)2
Things which are equal to the same thing are also equal to one another. Since
x2-300x+22500 = 12500 and
x2-300x+22500 = (x-150)2
then, according to the law of transitivity,
(x-150)2 = 12500
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-150)2 is
(x-150)2/2 =
(x-150)1 =
x-150
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-150 = √ 12500
Add 150 to both sides to obtain:
x = 150 + √ 12500
Since a square root has two values, one positive and the other negative
x2 - 300x + 10000 = 0
has two solutions:
x = 150 + √ 12500
or
x = 150 - √ 12500
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving x2-300x+10000 = 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 = -300
C = 10000
Accordingly, B2 - 4AC =
90000 - 40000 =
50000
Applying the quadratic formula :
300 ± √ 50000
x = ————————
2
Can √ 50000 be simplified ?
Yes! The prime factorization of 50000 is
2•2•2•2•5•5•5•5•5
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).
√ 50000 = √ 2•2•2•2•5•5•5•5•5 =2•2•5•5•√ 5 =
± 100 • √ 5
√ 5 , rounded to 4 decimal digits, is 2.2361
So now we are looking at:
x = ( 300 ± 100 • 2.236 ) / 2
Two real solutions:
x =(300+√50000)/2=150+50√ 5 = 261.803
or:
x =(300-√50000)/2=150-50√ 5 = 38.197
Two solutions were found :
- x =(300-√50000)/2=150-50√ 5 = 38.197
- x =(300+√50000)/2=150+50√ 5 = 261.803
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