Solution - Quadratic equations
Other Ways to Solve:
Step by Step Solution
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(32z2 - 10z) - 20 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 9z2-10z-20
The first term is, 9z2 its coefficient is 9 .
The middle term is, -10z its coefficient is -10 .
The last term, "the constant", is -20
Step-1 : Multiply the coefficient of the first term by the constant 9 • -20 = -180
Step-2 : Find two factors of -180 whose sum equals the coefficient of the middle term, which is -10 .
| -180 | + | 1 | = | -179 | ||
| -90 | + | 2 | = | -88 | ||
| -60 | + | 3 | = | -57 | ||
| -45 | + | 4 | = | -41 | ||
| -36 | + | 5 | = | -31 | ||
| -30 | + | 6 | = | -24 |
For tidiness, printing of 12 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 2 :
9z2 - 10z - 20 = 0
Step 3 :
Parabola, Finding the Vertex :
3.1 Find the Vertex of y = 9z2-10z-20
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, 9 , 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,Az2+Bz+C,the z -coordinate of the vertex is given by -B/(2A) . In our case the z coordinate is 0.5556
Plugging into the parabola formula 0.5556 for z we can calculate the y -coordinate :
y = 9.0 * 0.56 * 0.56 - 10.0 * 0.56 - 20.0
or y = -22.778
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 9z2-10z-20
Axis of Symmetry (dashed) {z}={ 0.56}
Vertex at {z,y} = { 0.56,-22.78}
z -Intercepts (Roots) :
Root 1 at {z,y} = {-1.04, 0.00}
Root 2 at {z,y} = { 2.15, 0.00}
Solve Quadratic Equation by Completing The Square
3.2 Solving 9z2-10z-20 = 0 by Completing The Square .
Divide both sides of the equation by 9 to have 1 as the coefficient of the first term :
z2-(10/9)z-(20/9) = 0
Add 20/9 to both side of the equation :
z2-(10/9)z = 20/9
Now the clever bit: Take the coefficient of z , which is 10/9 , divide by two, giving 5/9 , and finally square it giving 25/81
Add 25/81 to both sides of the equation :
On the right hand side we have :
20/9 + 25/81 The common denominator of the two fractions is 81 Adding (180/81)+(25/81) gives 205/81
So adding to both sides we finally get :
z2-(10/9)z+(25/81) = 205/81
Adding 25/81 has completed the left hand side into a perfect square :
z2-(10/9)z+(25/81) =
(z-(5/9)) • (z-(5/9)) =
(z-(5/9))2
Things which are equal to the same thing are also equal to one another. Since
z2-(10/9)z+(25/81) = 205/81 and
z2-(10/9)z+(25/81) = (z-(5/9))2
then, according to the law of transitivity,
(z-(5/9))2 = 205/81
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(z-(5/9))2 is
(z-(5/9))2/2 =
(z-(5/9))1 =
z-(5/9)
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
z-(5/9) = √ 205/81
Add 5/9 to both sides to obtain:
z = 5/9 + √ 205/81
Since a square root has two values, one positive and the other negative
z2 - (10/9)z - (20/9) = 0
has two solutions:
z = 5/9 + √ 205/81
or
z = 5/9 - √ 205/81
Note that √ 205/81 can be written as
√ 205 / √ 81 which is √ 205 / 9
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving 9z2-10z-20 = 0 by the Quadratic Formula .
According to the Quadratic Formula, z , the solution for Az2+Bz+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
z = ————————
2A
In our case, A = 9
B = -10
C = -20
Accordingly, B2 - 4AC =
100 - (-720) =
820
Applying the quadratic formula :
10 ± √ 820
z = —————
18
Can √ 820 be simplified ?
Yes! The prime factorization of 820 is
2•2•5•41
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).
√ 820 = √ 2•2•5•41 =
± 2 • √ 205
√ 205 , rounded to 4 decimal digits, is 14.3178
So now we are looking at:
z = ( 10 ± 2 • 14.318 ) / 18
Two real solutions:
z =(10+√820)/18=(5+√ 205 )/9= 2.146
or:
z =(10-√820)/18=(5-√ 205 )/9= -1.035
Two solutions were found :
- z =(10-√820)/18=(5-√ 205 )/9= -1.035
- z =(10+√820)/18=(5+√ 205 )/9= 2.146
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