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". 1 more similar replacement(s).
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((4 • (x4)) + 7x2) - 2 = 0Step 2 :
Equation at the end of step 2 :
(22x4 + 7x2) - 2 = 0
Step 3 :
Trying to factor by splitting the middle term
3.1 Factoring 4x4+7x2-2
The first term is, 4x4 its coefficient is 4 .
The middle term is, +7x2 its coefficient is 7 .
The last term, "the constant", is -2
Step-1 : Multiply the coefficient of the first term by the constant 4 • -2 = -8
Step-2 : Find two factors of -8 whose sum equals the coefficient of the middle term, which is 7 .
| -8 | + | 1 | = | -7 | ||
| -4 | + | 2 | = | -2 | ||
| -2 | + | 4 | = | 2 | ||
| -1 | + | 8 | = | 7 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 8
4x4 - 1x2 + 8x2 - 2
Step-4 : Add up the first 2 terms, pulling out like factors :
x2 • (4x2-1)
Add up the last 2 terms, pulling out common factors :
2 • (4x2-1)
Step-5 : Add up the four terms of step 4 :
(x2+2) • (4x2-1)
Which is the desired factorization
Trying to factor as a Difference of Squares :
3.2 Factoring: 4x2-1
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 : 4 is the square of 2
Check : 1 is the square of 1
Check : x2 is the square of x1
Factorization is : (2x + 1) • (2x - 1)
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = x2 + 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 | 3.00 | ||||||
| -2 | 1 | -2.00 | 6.00 | ||||||
| 1 | 1 | 1.00 | 3.00 | ||||||
| 2 | 1 | 2.00 | 6.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 3 :
(2x + 1) • (2x - 1) • (x2 + 2) = 0
Step 4 :
Theory - Roots of a product :
4.1 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 :
4.2 Solve : 2x+1 = 0
Subtract 1 from both sides of the equation :
2x = -1
Divide both sides of the equation by 2:
x = -1/2 = -0.500
Solving a Single Variable Equation :
4.3 Solve : 2x-1 = 0
Add 1 to both sides of the equation :
2x = 1
Divide both sides of the equation by 2:
x = 1/2 = 0.500
Solving a Single Variable Equation :
4.4 Solve : x2+2 = 0
Subtract 2 from both sides of the equation :
x2 = -2
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -2
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -2 =
√ -1• 2 =
√ -1 •√ 2 =
i • √ 2
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 1.4142 i
x= 0.0000 - 1.4142 i
Supplement : Solving Quadratic Equation Directly
Solving 4x4+7x2-2 = 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 :
5.1 Solve 4x4+7x2-2 = 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 :
4w2+7w-2 = 0
Solving this new equation using the quadratic formula we get two real solutions :
0.2500 or -2.0000
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
4x4+7x2-2 = 0
are either :
x =√ 0.250 = 0.50000 or :
x =√ 0.250 = -0.50000 or :
x =√-2.000 = 0.0 + 1.41421 i or :
x =√-2.000 = 0.0 - 1.41421 i
Four solutions were found :
- x= 0.0000 - 1.4142 i
- x= 0.0000 + 1.4142 i
- x = 1/2 = 0.500
- x = -1/2 = -0.500
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