Solution - Quadratic equations
Other Ways to Solve:
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x2" was replaced by "x^2".
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2+20x+360
The first term is, x2 its coefficient is 1 .
The middle term is, +20x its coefficient is 20 .
The last term, "the constant", is +360
Step-1 : Multiply the coefficient of the first term by the constant 1 • 360 = 360
Step-2 : Find two factors of 360 whose sum equals the coefficient of the middle term, which is 20 .
-360 | + | -1 | = | -361 | ||
-180 | + | -2 | = | -182 | ||
-120 | + | -3 | = | -123 | ||
-90 | + | -4 | = | -94 | ||
-72 | + | -5 | = | -77 | ||
-60 | + | -6 | = | -66 |
For tidiness, printing of 42 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 + 20x + 360 = 0
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2+20x+360
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 -10.0000
Plugging into the parabola formula -10.0000 for x we can calculate the y -coordinate :
y = 1.0 * -10.00 * -10.00 + 20.0 * -10.00 + 360.0
or y = 260.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2+20x+360
Axis of Symmetry (dashed) {x}={-10.00}
Vertex at {x,y} = {-10.00,260.00}
Function has no real roots
Solve Quadratic Equation by Completing The Square
2.2 Solving x2+20x+360 = 0 by Completing The Square .
Subtract 360 from both side of the equation :
x2+20x = -360
Now the clever bit: Take the coefficient of x , which is 20 , divide by two, giving 10 , and finally square it giving 100
Add 100 to both sides of the equation :
On the right hand side we have :
-360 + 100 or, (-360/1)+(100/1)
The common denominator of the two fractions is 1 Adding (-360/1)+(100/1) gives -260/1
So adding to both sides we finally get :
x2+20x+100 = -260
Adding 100 has completed the left hand side into a perfect square :
x2+20x+100 =
(x+10) • (x+10) =
(x+10)2
Things which are equal to the same thing are also equal to one another. Since
x2+20x+100 = -260 and
x2+20x+100 = (x+10)2
then, according to the law of transitivity,
(x+10)2 = -260
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+10)2 is
(x+10)2/2 =
(x+10)1 =
x+10
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x+10 = √ -260
Subtract 10 from both sides to obtain:
x = -10 + √ -260
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Since a square root has two values, one positive and the other negative
x2 + 20x + 360 = 0
has two solutions:
x = -10 + √ 260 • i
or
x = -10 - √ 260 • i
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving x2+20x+360 = 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 = 20
C = 360
Accordingly, B2 - 4AC =
400 - 1440 =
-1040
Applying the quadratic formula :
-20 ± √ -1040
x = ————————
2
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly,√ -1040 =
√ 1040 • (-1) =
√ 1040 • √ -1 =
± √ 1040 • i
Can √ 1040 be simplified ?
Yes! The prime factorization of 1040 is
2•2•2•2•5•13
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).
√ 1040 = √ 2•2•2•2•5•13 =2•2•√ 65 =
± 4 • √ 65
√ 65 , rounded to 4 decimal digits, is 8.0623
So now we are looking at:
x = ( -20 ± 4 • 8.062 i ) / 2
Two imaginary solutions :
x =(-20+√-1040)/2=-10+2i√ 65 = -10.0000+16.1245i or:
x =(-20-√-1040)/2=-10-2i√ 65 = -10.0000-16.1245i
Two solutions were found :
- x =(-20-√-1040)/2=-10-2i√ 65 = -10.0000-16.1245i
- x =(-20+√-1040)/2=-10+2i√ 65 = -10.0000+16.1245i
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