Solution - Equations reducible to quadratic form
Other Ways to Solve
Equations reducible to quadratic formStep by Step Solution
Step by step solution :
Step 1 :
Trying to factor as a Difference of Cubes:
1.1 Factoring: x27-216
Theory : A difference 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 : 216 is the cube of 6
Check : x27 is the cube of x9
Factorization is :
(x9 - 6) • (x18 + 6x9 + 36)
Trying to factor as a Difference of Cubes:
1.2 Factoring: x9 - 6
Check : 6 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 - 6
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 -6.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,6
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -7.00 | ||||||
| -2 | 1 | -2.00 | -518.00 | ||||||
| -3 | 1 | -3.00 | -19689.00 | ||||||
| -6 | 1 | -6.00 | -10077702.00 | ||||||
| 1 | 1 | 1.00 | -5.00 | ||||||
| 2 | 1 | 2.00 | 506.00 | ||||||
| 3 | 1 | 3.00 | 19677.00 | ||||||
| 6 | 1 | 6.00 | 10077690.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor by splitting the middle term
1.4 Factoring x18 + 6x9 + 36
The first term is, x18 its coefficient is 1 .
The middle term is, +6x9 its coefficient is 6 .
The last term, "the constant", is +36
Step-1 : Multiply the coefficient of the first term by the constant 1 • 36 = 36
Step-2 : Find two factors of 36 whose sum equals the coefficient of the middle term, which is 6 .
| -36 | + | -1 | = | -37 | ||
| -18 | + | -2 | = | -20 | ||
| -12 | + | -3 | = | -15 | ||
| -9 | + | -4 | = | -13 | ||
| -6 | + | -6 | = | -12 | ||
| -4 | + | -9 | = | -13 |
For tidiness, printing of 12 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 :
(x9 - 6) • (x18 + 6x9 + 36) = 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 : x9-6 = 0
Add 6 to both sides of the equation :
x9 = 6
x = 9th root of (6)
The equation has one real solution
This solution is x = 9th root of 6 = 1.2203
Solving a Single Variable Equation :
Equations which are reducible to quadratic :
2.3 Solve x18+6x9+36 = 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+6w+36 = 0
Solving this new equation using the quadratic formula we get two imaginary solutions :
w = -3.0000 ± 5.1962 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 -3.000 + 5.196 i are:
9th roots of -3.000- 5.196 i :
19 solutions were found :
x = 9th root of 6 = 1.2203How did we do?
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