Solution - Quadratic equations
Other Ways to Solve:
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring s2+70s-600
The first term is, s2 its coefficient is 1 .
The middle term is, +70s its coefficient is 70 .
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 70 .
-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 :
s2 + 70s - 600 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = s2+70s-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,As2+Bs+C,the s -coordinate of the vertex is given by -B/(2A) . In our case the s coordinate is -35.0000
Plugging into the parabola formula -35.0000 for s we can calculate the y -coordinate :
y = 1.0 * -35.00 * -35.00 + 70.0 * -35.00 - 600.0
or y = -1825.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = s2+70s-600
Axis of Symmetry (dashed) {s}={-35.00}
Vertex at {s,y} = {-35.00,-1825.00}
s -Intercepts (Roots) :
Root 1 at {s,y} = {-77.72, 0.00}
Root 2 at {s,y} = { 7.72, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving s2+70s-600 = 0 by Completing The Square .
Add 600 to both side of the equation :
s2+70s = 600
Now the clever bit: Take the coefficient of s , which is 70 , divide by two, giving 35 , and finally square it giving 1225
Add 1225 to both sides of the equation :
On the right hand side we have :
600 + 1225 or, (600/1)+(1225/1)
The common denominator of the two fractions is 1 Adding (600/1)+(1225/1) gives 1825/1
So adding to both sides we finally get :
s2+70s+1225 = 1825
Adding 1225 has completed the left hand side into a perfect square :
s2+70s+1225 =
(s+35) • (s+35) =
(s+35)2
Things which are equal to the same thing are also equal to one another. Since
s2+70s+1225 = 1825 and
s2+70s+1225 = (s+35)2
then, according to the law of transitivity,
(s+35)2 = 1825
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
(s+35)2 is
(s+35)2/2 =
(s+35)1 =
s+35
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
s+35 = √ 1825
Subtract 35 from both sides to obtain:
s = -35 + √ 1825
Since a square root has two values, one positive and the other negative
s2 + 70s - 600 = 0
has two solutions:
s = -35 + √ 1825
or
s = -35 - √ 1825
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving s2+70s-600 = 0 by the Quadratic Formula .
According to the Quadratic Formula, s , the solution for As2+Bs+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
s = ————————
2A
In our case, A = 1
B = 70
C = -600
Accordingly, B2 - 4AC =
4900 - (-2400) =
7300
Applying the quadratic formula :
-70 ± √ 7300
s = ———————
2
Can √ 7300 be simplified ?
Yes! The prime factorization of 7300 is
2•2•5•5•73
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).
√ 7300 = √ 2•2•5•5•73 =2•5•√ 73 =
± 10 • √ 73
√ 73 , rounded to 4 decimal digits, is 8.5440
So now we are looking at:
s = ( -70 ± 10 • 8.544 ) / 2
Two real solutions:
s =(-70+√7300)/2=-35+5√ 73 = 7.720
or:
s =(-70-√7300)/2=-35-5√ 73 = -77.720
Two solutions were found :
- s =(-70-√7300)/2=-35-5√ 73 = -77.720
- s =(-70+√7300)/2=-35+5√ 73 = 7.720
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