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. 


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 11 because it is a perfect square.


The above method have given two distinct values:

βˆ’1β€…andβ€…1-1 \> and \> 1  and are not both the roots of quadratic function xΒ²βˆ’2x+1xΒ²-2x+1 


When x=βˆ’1x=-1 is substituted in the quadratic equation

(βˆ’1)Β²βˆ’2(βˆ’1)+1β‰ 0(-1)Β²-2(-1)+1\ne0


The method is assumed to be same as x(xβˆ’2)=0.β€…x(x-2)=0. \>

In these case either x=0β€…orβ€…xβˆ’2=0x=0\>or\>x-2=0

This will always be true because:


For Ξ±β€…andβ€…Ξ²β€…βˆˆβ€…R\alpha\>and \>\beta\>\in\>\Reals

when either is zero, then

Ξ±Ξ²=O\alpha\beta=O



But for  x(xβˆ’2)=βˆ’1x(x-2)=-1 and x=βˆ’1x=-1the equation can only be true if 

xβˆ’2=1.x-2=1.

Otherwise this method will give invalid roots.

The method also treat the function

f(x)=x(xβˆ’2)β€…f(x) =x(x-2) \>

as a function with two variables.

In that for forβ€…f(x)=βˆ’1for \>f(x)=-1

the first x=βˆ’1x=-1 and (xβˆ’2)=1(x-2)=1

The second xx in the expression becomes 33 so that x(xβˆ’2)=βˆ’1x(x-2)=-1

The method is invalid







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