Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
x^(3)+(x)/(2)-(9)=0
Step by step solution :
Step 1 :
x
Simplify —
2
Equation at the end of step 1 :
x ((x3) + —) - 9 = 0 2Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Adding a fraction to a whole
Rewrite the whole as a fraction using 2 as the denominator :
x3 x3 • 2
x3 = —— = ——————
1 2
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
x3 • 2 + x 2x3 + x
—————————— = ———————
2 2
Equation at the end of step 2 :
(2x3 + x)
————————— - 9 = 0
2
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 2 as the denominator :
9 9 • 2
9 = — = —————
1 2
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
2x3 + x = x • (2x2 + 1)
Polynomial Roots Calculator :
4.2 Find roots (zeroes) of : F(x) = 2x2 + 1
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 2 and the Trailing Constant is 1.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 3.00 | ||||||
-1 | 2 | -0.50 | 1.50 | ||||||
1 | 1 | 1.00 | 3.00 | ||||||
1 | 2 | 0.50 | 1.50 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
4.3 Adding up the two equivalent fractions
x • (2x2+1) - (9 • 2) 2x3 + x - 18
————————————————————— = ————————————
2 2
Polynomial Roots Calculator :
4.4 Find roots (zeroes) of : F(x) = 2x3 + x - 18
See theory in step 4.2
In this case, the Leading Coefficient is 2 and the Trailing Constant is -18.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,2 ,3 ,6 ,9 ,18
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | -21.00 | ||||||
-1 | 2 | -0.50 | -18.75 | ||||||
-2 | 1 | -2.00 | -36.00 | ||||||
-3 | 1 | -3.00 | -75.00 | ||||||
-3 | 2 | -1.50 | -26.25 | ||||||
-6 | 1 | -6.00 | -456.00 | ||||||
-9 | 1 | -9.00 | -1485.00 | ||||||
-9 | 2 | -4.50 | -204.75 | ||||||
-18 | 1 | -18.00 | -11700.00 | ||||||
1 | 1 | 1.00 | -15.00 | ||||||
1 | 2 | 0.50 | -17.25 | ||||||
2 | 1 | 2.00 | 0.00 | x - 2 | |||||
3 | 1 | 3.00 | 39.00 | ||||||
3 | 2 | 1.50 | -9.75 | ||||||
6 | 1 | 6.00 | 420.00 | ||||||
9 | 1 | 9.00 | 1449.00 | ||||||
9 | 2 | 4.50 | 168.75 | ||||||
18 | 1 | 18.00 | 11664.00 |
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms
In our case this means that
2x3 + x - 18
can be divided with x - 2
Polynomial Long Division :
4.5 Polynomial Long Division
Dividing : 2x3 + x - 18
("Dividend")
By : x - 2 ("Divisor")
dividend | 2x3 | + | x | - | 18 | ||||
- divisor | * 2x2 | 2x3 | - | 4x2 | |||||
remainder | 4x2 | + | x | - | 18 | ||||
- divisor | * 4x1 | 4x2 | - | 8x | |||||
remainder | 9x | - | 18 | ||||||
- divisor | * 9x0 | 9x | - | 18 | |||||
remainder | 0 |
Quotient : 2x2+4x+9 Remainder: 0
Trying to factor by splitting the middle term
4.6 Factoring 2x2+4x+9
The first term is, 2x2 its coefficient is 2 .
The middle term is, +4x its coefficient is 4 .
The last term, "the constant", is +9
Step-1 : Multiply the coefficient of the first term by the constant 2 • 9 = 18
Step-2 : Find two factors of 18 whose sum equals the coefficient of the middle term, which is 4 .
-18 | + | -1 | = | -19 | ||
-9 | + | -2 | = | -11 | ||
-6 | + | -3 | = | -9 | ||
-3 | + | -6 | = | -9 | ||
-2 | + | -9 | = | -11 | ||
-1 | + | -18 | = | -19 | ||
1 | + | 18 | = | 19 | ||
2 | + | 9 | = | 11 | ||
3 | + | 6 | = | 9 | ||
6 | + | 3 | = | 9 | ||
9 | + | 2 | = | 11 | ||
18 | + | 1 | = | 19 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 4 :
(2x2 + 4x + 9) • (x - 2)
———————————————————————— = 0
2
Step 5 :
When a fraction equals zero :
5.1 When a fraction equals zero ...
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
(2x2+4x+9)•(x-2)
———————————————— • 2 = 0 • 2
2
Now, on the left hand side, the 2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
(2x2+4x+9) • (x-2) = 0
Theory - Roots of a product :
5.2 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.
Parabola, Finding the Vertex :
5.3 Find the Vertex of y = 2x2+4x+9
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, 2 , 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 -1.0000
Plugging into the parabola formula -1.0000 for x we can calculate the y -coordinate :
y = 2.0 * -1.00 * -1.00 + 4.0 * -1.00 + 9.0
or y = 7.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 2x2+4x+9
Axis of Symmetry (dashed) {x}={-1.00}
Vertex at {x,y} = {-1.00, 7.00}
Function has no real roots
Solve Quadratic Equation by Completing The Square
5.4 Solving 2x2+4x+9 = 0 by Completing The Square .
Divide both sides of the equation by 2 to have 1 as the coefficient of the first term :
x2+2x+(9/2) = 0
Subtract 9/2 from both side of the equation :
x2+2x = -9/2
Now the clever bit: Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1
Add 1 to both sides of the equation :
On the right hand side we have :
-9/2 + 1 or, (-9/2)+(1/1)
The common denominator of the two fractions is 2 Adding (-9/2)+(2/2) gives -7/2
So adding to both sides we finally get :
x2+2x+1 = -7/2
Adding 1 has completed the left hand side into a perfect square :
x2+2x+1 =
(x+1) • (x+1) =
(x+1)2
Things which are equal to the same thing are also equal to one another. Since
x2+2x+1 = -7/2 and
x2+2x+1 = (x+1)2
then, according to the law of transitivity,
(x+1)2 = -7/2
We'll refer to this Equation as Eq. #5.4.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x+1)2 is
(x+1)2/2 =
(x+1)1 =
x+1
Now, applying the Square Root Principle to Eq. #5.4.1 we get:
x+1 = √ -7/2
Subtract 1 from both sides to obtain:
x = -1 + √ -7/2
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 + 2x + (9/2) = 0
has two solutions:
x = -1 + √ 7/2 • i
or
x = -1 - √ 7/2 • i
Note that √ 7/2 can be written as
√ 7 / √ 2
Solve Quadratic Equation using the Quadratic Formula
5.5 Solving 2x2+4x+9 = 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 = 2
B = 4
C = 9
Accordingly, B2 - 4AC =
16 - 72 =
-56
Applying the quadratic formula :
-4 ± √ -56
x = ——————
4
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,√ -56 =
√ 56 • (-1) =
√ 56 • √ -1 =
± √ 56 • i
Can √ 56 be simplified ?
Yes! The prime factorization of 56 is
2•2•2•7
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).
√ 56 = √ 2•2•2•7 =
± 2 • √ 14
√ 14 , rounded to 4 decimal digits, is 3.7417
So now we are looking at:
x = ( -4 ± 2 • 3.742 i ) / 4
Two imaginary solutions :
x =(-4+√-56)/4=-1+i/2√ 14 = -1.0000+1.8708i or:
x =(-4-√-56)/4=-1-i/2√ 14 = -1.0000-1.8708i
Solving a Single Variable Equation :
5.6 Solve : x-2 = 0
Add 2 to both sides of the equation :
x = 2
Three solutions were found :
- x = 2
- x =(-4-√-56)/4=-1-i/2√ 14 = -1.0000-1.8708i
- x =(-4+√-56)/4=-1+i/2√ 14 = -1.0000+1.8708i
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