Solution - Equations reducible to quadratic form
Other Ways to Solve
Equations reducible to quadratic formStep 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 :
Equation at the end of step 1 :
((x4) + 32x2) - 10 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring x4+9x2-10
The first term is, x4 its coefficient is 1 .
The middle term is, +9x2 its coefficient is 9 .
The last term, "the constant", is -10
Step-1 : Multiply the coefficient of the first term by the constant 1 • -10 = -10
Step-2 : Find two factors of -10 whose sum equals the coefficient of the middle term, which is 9 .
-10 | + | 1 | = | -9 | ||
-5 | + | 2 | = | -3 | ||
-2 | + | 5 | = | 3 | ||
-1 | + | 10 | = | 9 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 10
x4 - 1x2 + 10x2 - 10
Step-4 : Add up the first 2 terms, pulling out like factors :
x2 • (x2-1)
Add up the last 2 terms, pulling out common factors :
10 • (x2-1)
Step-5 : Add up the four terms of step 4 :
(x2+10) • (x2-1)
Which is the desired factorization
Polynomial Roots Calculator :
2.2 Find roots (zeroes) of : F(x) = x2+10
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 10.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,5 ,10
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 11.00 | ||||||
-2 | 1 | -2.00 | 14.00 | ||||||
-5 | 1 | -5.00 | 35.00 | ||||||
-10 | 1 | -10.00 | 110.00 | ||||||
1 | 1 | 1.00 | 11.00 | ||||||
2 | 1 | 2.00 | 14.00 | ||||||
5 | 1 | 5.00 | 35.00 | ||||||
10 | 1 | 10.00 | 110.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor as a Difference of Squares :
2.3 Factoring: x2-1
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 1 is the square of 1
Check : x2 is the square of x1
Factorization is : (x + 1) • (x - 1)
Equation at the end of step 2 :
(x2 + 10) • (x + 1) • (x - 1) = 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 : x2+10 = 0
Subtract 10 from both sides of the equation :
x2 = -10
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -10
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -10 =
√ -1• 10 =
√ -1 •√ 10 =
i • √ 10
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 3.1623 i
x= 0.0000 - 3.1623 i
Solving a Single Variable Equation :
3.3 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Solving a Single Variable Equation :
3.4 Solve : x-1 = 0
Add 1 to both sides of the equation :
x = 1
Supplement : Solving Quadratic Equation Directly
Solving x4+9x2-10 = 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
Solving a Single Variable Equation :
Equations which are reducible to quadratic :
4.1 Solve x4+9x2-10 = 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 = x2 transforms the equation into :
w2+9w-10 = 0
Solving this new equation using the quadratic formula we get two real solutions :
1.0000 or -10.0000
Now that we know the value(s) of w , we can calculate x since x is √ w
Doing just this we discover that the solutions of
x4+9x2-10 = 0
are either :
x =√ 1.000 = 1.00000 or :
x =√ 1.000 = -1.00000 or :
x =√-10.000 = 0.0 + 3.16228 i or :
x =√-10.000 = 0.0 - 3.16228 i
Four solutions were found :
- x = 1
- x = -1
- x= 0.0000 - 3.1623 i
- x= 0.0000 + 3.1623 i
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