Examine the following solution to π₯ 2 β 2π₯ = β1: π₯(π₯ β 2) = β1 π₯ = β1 or π₯ β 2 = β1 π₯ = β1 or π₯ = 1 Is this method correct? Explain.Β
QUESTION
Examine the following solution to xΒ²β 2x = β1:
x(x β 2) = β1
x = β1
or x β 2 = β1
x = β1 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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