Solution - Equations reducible to quadratic form
Other Ways to Solve
Equations reducible to quadratic formStep 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^26*x-(-8)=0
Step by step solution :
Step 1 :
Trying to factor as a Sum of Cubes :
1.1 Factoring: x27+8
Theory : A sum of two perfect cubes, a3 + b3 can be factored into :
(a+b) • (a2-ab+b2)
Proof : (a+b) • (a2-ab+b2) =
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+b3
Check : 8 is the cube of 2
Check : x27 is the cube of x9
Factorization is :
(x9 + 2) • (x18 - 2x9 + 4)
Trying to factor as a Sum of Cubes :
1.2 Factoring: x9 + 2
Check : 2 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Polynomial Roots Calculator :
1.3 Find roots (zeroes) of : F(x) = x9 + 2
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 1 and the Trailing Constant is 2.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 1.00 | ||||||
| -2 | 1 | -2.00 | -510.00 | ||||||
| 1 | 1 | 1.00 | 3.00 | ||||||
| 2 | 1 | 2.00 | 514.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor by splitting the middle term
1.4 Factoring x18 - 2x9 + 4
The first term is, x18 its coefficient is 1 .
The middle term is, -2x9 its coefficient is -2 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4
Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -2 .
| -4 | + | -1 | = | -5 | ||
| -2 | + | -2 | = | -4 | ||
| -1 | + | -4 | = | -5 | ||
| 1 | + | 4 | = | 5 | ||
| 2 | + | 2 | = | 4 | ||
| 4 | + | 1 | = | 5 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 1 :
(x9 + 2) • (x18 - 2x9 + 4) = 0
Step 2 :
Theory - Roots of a product :
2.1 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 :
2.2 Solve : x9+2 = 0
Subtract 2 from both sides of the equation :
x9 = -2
x = 9th root of (-2)
Negative numbers have real 9th roots.
9th root of (-2) = 9√ -1• 2 = 9√ -1 • 9√ 2 =(-1)•9√ 2
The equation has one real solution, a negative number This solution is x = negative 9th root of 2 = -1.0801
Solving a Single Variable Equation :
Equations which are reducible to quadratic :
2.3 Solve x18-2x9+4 = 0
This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x9 transforms the equation into :
w2-2w+4 = 0
Solving this new equation using the quadratic formula we get two imaginary solutions :
w = 1.0000 ± 1.7321 i
Now that we know the value(s) of w , we can calculate x since x is the 9 root of w
Since we are speaking 9th root, each of the two imaginary solutions of has 9 roots
Tiger finds these roots using de Moivre's Formula
The 9th roots of 1.000 + 1.732 i are:
9th roots of 1.000- 1.732 i :
19 solutions were found :
x = negative 9th root of 2 = -1.0801How did we do?
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