Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-190x-600
The first term is, x2 its coefficient is 1 .
The middle term is, -190x its coefficient is -190 .
The last term, "the constant", is -600
Step-1 : Multiply the coefficient of the first term by the constant 1 • -600 = -600
Step-2 : Find two factors of -600 whose sum equals the coefficient of the middle term, which is -190 .
| -600 | + | 1 | = | -599 | ||
| -300 | + | 2 | = | -298 | ||
| -200 | + | 3 | = | -197 | ||
| -150 | + | 4 | = | -146 | ||
| -120 | + | 5 | = | -115 | ||
| -100 | + | 6 | = | -94 |
For tidiness, printing of 18 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 - 190x - 600 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2-190x-600
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 95.0000
Plugging into the parabola formula 95.0000 for x we can calculate the y -coordinate :
y = 1.0 * 95.00 * 95.00 - 190.0 * 95.00 - 600.0
or y = -9625.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-190x-600
Axis of Symmetry (dashed) {x}={95.00}
Vertex at {x,y} = {95.00,-9625.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-3.11, 0.00}
Root 2 at {x,y} = {193.11, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving x2-190x-600 = 0 by Completing The Square .
Add 600 to both side of the equation :
x2-190x = 600
Now the clever bit: Take the coefficient of x , which is 190 , divide by two, giving 95 , and finally square it giving 9025
Add 9025 to both sides of the equation :
On the right hand side we have :
600 + 9025 or, (600/1)+(9025/1)
The common denominator of the two fractions is 1 Adding (600/1)+(9025/1) gives 9625/1
So adding to both sides we finally get :
x2-190x+9025 = 9625
Adding 9025 has completed the left hand side into a perfect square :
x2-190x+9025 =
(x-95) • (x-95) =
(x-95)2
Things which are equal to the same thing are also equal to one another. Since
x2-190x+9025 = 9625 and
x2-190x+9025 = (x-95)2
then, according to the law of transitivity,
(x-95)2 = 9625
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-95)2 is
(x-95)2/2 =
(x-95)1 =
x-95
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-95 = √ 9625
Add 95 to both sides to obtain:
x = 95 + √ 9625
Since a square root has two values, one positive and the other negative
x2 - 190x - 600 = 0
has two solutions:
x = 95 + √ 9625
or
x = 95 - √ 9625
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving x2-190x-600 = 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 = -190
C = -600
Accordingly, B2 - 4AC =
36100 - (-2400) =
38500
Applying the quadratic formula :
190 ± √ 38500
x = ————————
2
Can √ 38500 be simplified ?
Yes! The prime factorization of 38500 is
2•2•5•5•5•7•11
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).
√ 38500 = √ 2•2•5•5•5•7•11 =2•5•√ 385 =
± 10 • √ 385
√ 385 , rounded to 4 decimal digits, is 19.6214
So now we are looking at:
x = ( 190 ± 10 • 19.621 ) / 2
Two real solutions:
x =(190+√38500)/2=95+5√ 385 = 193.107
or:
x =(190-√38500)/2=95-5√ 385 = -3.107
Two solutions were found :
- x =(190-√38500)/2=95-5√ 385 = -3.107
- x =(190+√38500)/2=95+5√ 385 = 193.107
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