Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((2•7x2) - 23x) - 30 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 14x2-23x-30
The first term is, 14x2 its coefficient is 14 .
The middle term is, -23x its coefficient is -23 .
The last term, "the constant", is -30
Step-1 : Multiply the coefficient of the first term by the constant 14 • -30 = -420
Step-2 : Find two factors of -420 whose sum equals the coefficient of the middle term, which is -23 .
-420 | + | 1 | = | -419 | ||
-210 | + | 2 | = | -208 | ||
-140 | + | 3 | = | -137 | ||
-105 | + | 4 | = | -101 | ||
-84 | + | 5 | = | -79 | ||
-70 | + | 6 | = | -64 | ||
-60 | + | 7 | = | -53 | ||
-42 | + | 10 | = | -32 | ||
-35 | + | 12 | = | -23 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -35 and 12
14x2 - 35x + 12x - 30
Step-4 : Add up the first 2 terms, pulling out like factors :
7x • (2x-5)
Add up the last 2 terms, pulling out common factors :
6 • (2x-5)
Step-5 : Add up the four terms of step 4 :
(7x+6) • (2x-5)
Which is the desired factorization
Equation at the end of step 2 :
(2x - 5) • (7x + 6) = 0
Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
3.2 Solve : 2x-5 = 0
Add 5 to both sides of the equation :
2x = 5
Divide both sides of the equation by 2:
x = 5/2 = 2.500
Solving a Single Variable Equation :
3.3 Solve : 7x+6 = 0
Subtract 6 from both sides of the equation :
7x = -6
Divide both sides of the equation by 7:
x = -6/7 = -0.857
Supplement : Solving Quadratic Equation Directly
Solving 14x2-23x-30 = 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 = 14x2-23x-30
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, 14 , 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 0.8214
Plugging into the parabola formula 0.8214 for x we can calculate the y -coordinate :
y = 14.0 * 0.82 * 0.82 - 23.0 * 0.82 - 30.0
or y = -39.446
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 14x2-23x-30
Axis of Symmetry (dashed) {x}={ 0.82}
Vertex at {x,y} = { 0.82,-39.45}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-0.86, 0.00}
Root 2 at {x,y} = { 2.50, 0.00}
Solve Quadratic Equation by Completing The Square
4.2 Solving 14x2-23x-30 = 0 by Completing The Square .
Divide both sides of the equation by 14 to have 1 as the coefficient of the first term :
x2-(23/14)x-(15/7) = 0
Add 15/7 to both side of the equation :
x2-(23/14)x = 15/7
Now the clever bit: Take the coefficient of x , which is 23/14 , divide by two, giving 23/28 , and finally square it giving 529/784
Add 529/784 to both sides of the equation :
On the right hand side we have :
15/7 + 529/784 The common denominator of the two fractions is 784 Adding (1680/784)+(529/784) gives 2209/784
So adding to both sides we finally get :
x2-(23/14)x+(529/784) = 2209/784
Adding 529/784 has completed the left hand side into a perfect square :
x2-(23/14)x+(529/784) =
(x-(23/28)) • (x-(23/28)) =
(x-(23/28))2
Things which are equal to the same thing are also equal to one another. Since
x2-(23/14)x+(529/784) = 2209/784 and
x2-(23/14)x+(529/784) = (x-(23/28))2
then, according to the law of transitivity,
(x-(23/28))2 = 2209/784
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
(x-(23/28))2 is
(x-(23/28))2/2 =
(x-(23/28))1 =
x-(23/28)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
x-(23/28) = √ 2209/784
Add 23/28 to both sides to obtain:
x = 23/28 + √ 2209/784
Since a square root has two values, one positive and the other negative
x2 - (23/14)x - (15/7) = 0
has two solutions:
x = 23/28 + √ 2209/784
or
x = 23/28 - √ 2209/784
Note that √ 2209/784 can be written as
√ 2209 / √ 784 which is 47 / 28
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving 14x2-23x-30 = 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 = 14
B = -23
C = -30
Accordingly, B2 - 4AC =
529 - (-1680) =
2209
Applying the quadratic formula :
23 ± √ 2209
x = ——————
28
Can √ 2209 be simplified ?
Yes! The prime factorization of 2209 is
47•47
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).
√ 2209 = √ 47•47 =
± 47 • √ 1 =
± 47
So now we are looking at:
x = ( 23 ± 47) / 28
Two real solutions:
x =(23+√2209)/28=(23+47)/28= 2.500
or:
x =(23-√2209)/28=(23-47)/28= -0.857
Two solutions were found :
- x = -6/7 = -0.857
- x = 5/2 = 2.500
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