Question

Examine the following solution to 𝑥 2 − 2𝑥 = −1: 𝑥(𝑥 − 2) = −1 𝑥 = −1 or 𝑥 − 2 = −1 𝑥 = −1 or 𝑥 = 1 Is this method correct? Explain. 

Answer 1

QUESTION

Examine the following solution to x²− 2x = −1:

x(x − 2) = −1

x = −1

or x − 2 = −1

x = −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:

$-1 \> and \> 1$  and are not both the roots of quadratic function $x²-2x+1$ 

When $x=-1$ is substituted in the quadratic equation

$(-1)²-2(-1)+1\ne0$

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

In these case either $x=0\>or\>x-2=0$

This will always be true because:

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

when either is zero, then

$\alpha\beta=O$

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

$x-2=1.$

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