Solution - Linear equations with one unknown
Step by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
9*w^2+25-(30*w)=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(32w2 + 25) - 30w = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 9w2-30w+25
The first term is, 9w2 its coefficient is 9 .
The middle term is, -30w its coefficient is -30 .
The last term, "the constant", is +25
Step-1 : Multiply the coefficient of the first term by the constant 9 • 25 = 225
Step-2 : Find two factors of 225 whose sum equals the coefficient of the middle term, which is -30 .
| -225 | + | -1 | = | -226 | ||
| -75 | + | -3 | = | -78 | ||
| -45 | + | -5 | = | -50 | ||
| -25 | + | -9 | = | -34 | ||
| -15 | + | -15 | = | -30 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -15 and -15
9w2 - 15w - 15w - 25
Step-4 : Add up the first 2 terms, pulling out like factors :
3w • (3w-5)
Add up the last 2 terms, pulling out common factors :
5 • (3w-5)
Step-5 : Add up the four terms of step 4 :
(3w-5) • (3w-5)
Which is the desired factorization
Multiplying Exponential Expressions :
2.2 Multiply (3w-5) by (3w-5)
The rule says : To multiply exponential expressions which have the same base, add up their exponents.
In our case, the common base is (3w-5) and the exponents are :
1 , as (3w-5) is the same number as (3w-5)1
and 1 , as (3w-5) is the same number as (3w-5)1
The product is therefore, (3w-5)(1+1) = (3w-5)2
Equation at the end of step 2 :
(3w - 5)2 = 0
Step 3 :
Solving a Single Variable Equation :
3.1 Solve : (3w-5)2 = 0
(3w-5) 2 represents, in effect, a product of 2 terms which is equal to zero
For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means : 3w-5 = 0
Add 5 to both sides of the equation :
3w = 5
Divide both sides of the equation by 3:
w = 5/3 = 1.667
Supplement : Solving Quadratic Equation Directly
Solving 9w2-30w+25 = 0 directly Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex :
4.1 Find the Vertex of y = 9w2-30w+25
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,Aw2+Bw+C,the w -coordinate of the vertex is given by -B/(2A) . In our case the w coordinate is 1.6667
Plugging into the parabola formula 1.6667 for w we can calculate the y -coordinate :
y = 9.0 * 1.67 * 1.67 - 30.0 * 1.67 + 25.0
or y = 0.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 9w2-30w+25
Vertex at {w,y} = { 1.67, 0.00}
w-Intercept (Root) :
One Root at {w,y}={ 1.67, 0.00}
Note that the root coincides with
the Vertex and the Axis of Symmetry
coinsides with the line w = 0
Solve Quadratic Equation by Completing The Square
4.2 Solving 9w2-30w+25 = 0 by Completing The Square .
Divide both sides of the equation by 9 to have 1 as the coefficient of the first term :
w2-(10/3)w+(25/9) = 0
Subtract 25/9 from both side of the equation :
w2-(10/3)w = -25/9
Now the clever bit: Take the coefficient of w , which is 10/3 , divide by two, giving 5/3 , and finally square it giving 25/9
Add 25/9 to both sides of the equation :
On the right hand side we have :
-25/9 + 25/9 The common denominator of the two fractions is 9 Adding (-25/9)+(25/9) gives 0/9
So adding to both sides we finally get :
w2-(10/3)w+(25/9) = 0
Adding 25/9 has completed the left hand side into a perfect square :
w2-(10/3)w+(25/9) =
(w-(5/3)) • (w-(5/3)) =
(w-(5/3))2
Things which are equal to the same thing are also equal to one another. Since
w2-(10/3)w+(25/9) = 0 and
w2-(10/3)w+(25/9) = (w-(5/3))2
then, according to the law of transitivity,
(w-(5/3))2 = 0
We'll refer to this Equation as Eq. #4.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(w-(5/3))2 is
(w-(5/3))2/2 =
(w-(5/3))1 =
w-(5/3)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
w-(5/3) = √ 0
Add 5/3 to both sides to obtain:
w = 5/3 + √ 0
The square root of zero is zero
This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.
The solution is:
w = 5/3
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving 9w2-30w+25 = 0 by the Quadratic Formula .
According to the Quadratic Formula, w , the solution for Aw2+Bw+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
w = ————————
2A
In our case, A = 9
B = -30
C = 25
Accordingly, B2 - 4AC =
900 - 900 =
0
Applying the quadratic formula :
30 ± √ 0
w = —————
18
The square root of zero is zero
This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.
The solution is:
w = 30 / 18 = 1.6667(rounded to 4 decimal places)
One solution was found :
w = 5/3 = 1.667How did we do?
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