Can you solve $(x - 2) (x - 3) = 5$ $(x-2)(x-3)=5$ by solving $x - 2 = 5 and x + 3 = 5$ $x-2=5 and x+3=5$ ?

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Answer 1 · 2018-03-17T17:34:23

No, this only works when the right side of the equation is $0$ $0$ .

Then if one of the terms on the left is equal to $0$ $0$ then $0 \times$ $0 xx$ anything is equal to $0$ $0$

The same is not true when one of the terms is equal to $5$ $5$

Can you solve (x−2)(x−3)=5(x-2)(x-3)=5 by solving x−2=5andx+3=5x-2=5 and x+3=5?