Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): ".2" was replaced by "(2/10)".
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
f*(2/10)*f+4-(8)=0
Step by step solution :
Step 1 :
1
Simplify —
5
Equation at the end of step 1 :
1 (((f • —) • f) + 4) - 8 = 0 5Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Adding a whole to a fraction
Rewrite the whole as a fraction using 5 as the denominator :
4 4 • 5
4 = — = —————
1 5
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:
f2 + 4 • 5 f2 + 20
—————————— = ———————
5 5
Equation at the end of step 2 :
(f2 + 20)
————————— - 8 = 0
5
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 5 as the denominator :
8 8 • 5
8 = — = —————
1 5
Polynomial Roots Calculator :
3.2 Find roots (zeroes) of : F(f) = f2 + 20
Polynomial Roots Calculator is a set of methods aimed at finding values of f for which F(f)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers f 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 20.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 21.00 | ||||||
| -2 | 1 | -2.00 | 24.00 | ||||||
| -4 | 1 | -4.00 | 36.00 | ||||||
| -5 | 1 | -5.00 | 45.00 | ||||||
| -10 | 1 | -10.00 | 120.00 | ||||||
| -20 | 1 | -20.00 | 420.00 | ||||||
| 1 | 1 | 1.00 | 21.00 | ||||||
| 2 | 1 | 2.00 | 24.00 | ||||||
| 4 | 1 | 4.00 | 36.00 | ||||||
| 5 | 1 | 5.00 | 45.00 | ||||||
| 10 | 1 | 10.00 | 120.00 | ||||||
| 20 | 1 | 20.00 | 420.00 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
3.3 Adding up the two equivalent fractions
(f2+20) - (8 • 5) f2 - 20
————————————————— = ———————
5 5
Trying to factor as a Difference of Squares :
3.4 Factoring: f2 - 20
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 : 20 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Equation at the end of step 3 :
f2 - 20
——————— = 0
5
Step 4 :
When a fraction equals zero :
4.1 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:
f2-20
————— • 5 = 0 • 5
5
Now, on the left hand side, the 5 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
f2-20 = 0
Solving a Single Variable Equation :
4.2 Solve : f2-20 = 0
Add 20 to both sides of the equation :
f2 = 20
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
f = ± √ 20
Can √ 20 be simplified ?
Yes! The prime factorization of 20 is
2•2•5
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).
√ 20 = √ 2•2•5 =
± 2 • √ 5
The equation has two real solutions
These solutions are f = 2 • ± √5 = ± 4.4721
Two solutions were found :
f = 2 • ± √5 = ± 4.4721How did we do?
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