Question

(3-x)^5+(x-5)^5=-32
what are the roots of this equation​

Answer 1

Answer:

One solution was found :

x = 5

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(3-5)^5+(x-5)^5-(-32)=0

Step by step solution :

Step 1 :

1.1

Negative number raised to an odd power is negative

To show this we begin with (-7)5 ,a negative number raised to an odd exponent:

(-7)5 can be written as (-7)•(-7)•(-7)•(-7)•(-7)

Now, using the rule that says minus times minus is plus, (-7)5 can be written as (49)•(49)•(-7) which is a negative number.

We showed that (-7)5 is a negative number

Using the same arguments as above, replacing (-7) by any negative number, and replacing the exponent 5 by any odd exponent, we proved which had to be proved

Equation at the end of step 1 :

( -25 + (x - 5)5) - -32 = 0

Step 2 :

Trying to factor by pulling out :

2.1 Factoring: x5-25x4+250x3-1250x2+3125x-3125

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: 3125x-3125

Group 2: 250x3-1250x2

Group 3: x5-25x4

Pull out from each group separately :

Group 1: (x-1) • (3125)

Group 2: (x-5) • (250x2)

Group 3: (x-25) • (x4)

Looking for common sub-expressions :

Group 1: (x-1) • (3125)

Group 3: (x-25) • (x4)

Group 2: (x-5) • (250x2)

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