Answer to Question #262615 in Algebra for Amani

Question #262615

Examine the following solution to π‘₯ 2 βˆ’ 2π‘₯ = βˆ’1: π‘₯(π‘₯ βˆ’ 2) = βˆ’1 π‘₯ = βˆ’1 or π‘₯ βˆ’ 2 = βˆ’1 π‘₯ = βˆ’1 or π‘₯ = 1 Is this method correct? Explain.Β 


1
Expert's answer
2021-11-09T16:53:23-0500

QUESTION


Examine the following solution to xΒ²βˆ’ 2x = βˆ’1:

x(x βˆ’ 2) = βˆ’1

x = βˆ’1

or x βˆ’ 2 = βˆ’1

x = βˆ’1 or x = 1

is this method correct? explain.


ANSWER


The method is not correct

The root of the quadratic equation is "1" because it is a perfect square.


The above method have given two distinct values:

"-1 \\> and \\> 1"Β  and are not both the roots of quadratic functionΒ "x\u00b2-2x+1"Β 


WhenΒ "x=-1"Β is substituted in the quadratic equation

"(-1)\u00b2-2(-1)+1\\ne0"


The method is assumed to be same as "x(x-2)=0. \\>"

In these case eitherΒ "x=0\\>or\\>x-2=0"

This will always be true because:


For "\\alpha\\>and \\>\\beta\\>\\in\\>\\Reals"

when either is zero, then

"\\alpha\\beta=O"



But for Β "x(x-2)=-1"Β andΒ "x=-1"the equation can only be true ifΒ 

"x-2=1."

Otherwise this method will give invalid roots.

The method also treat the function

"f(x) =x(x-2) \\>"

as a function with two variables.

In that for "for \\>f(x)=-1"

the first "x=-1" and "(x-2)=1"

The second "x" in the expression becomes "3" so that "x(x-2)=-1"

The method is invalid







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