Algebra what is the actual answer ?

Mar 2019
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$$(a-b)^2=a^2-b^2$$

Which values of a and b show that this is not true?
 
Oct 2018
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$(a-b)^{2} = a^{2} -2ab + b^{2} = a^{2}-b^{2}$

Subtract $a^{2}$ both sides

$-2ab = -2b^{2}$

$a=b$

So just choose any $a$ and $b$ such that $a \neq b$ (edit) and $b \neq 0$ as Romsek mentioned
 
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romsek

Math Team
Sep 2015
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$(a-b)^2 = a^2 - 2ab + b^2$

$a^2 - 2ab + b^2 = a^2 - b^2$

$2b^2 = 2ab$

$\text{If $b = 0$ this is true $\forall a$}$

$\text{otherwise $b=a$}$

$\text{So the original equation is only true if $b=0 \vee a=b$}$
 
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Mar 2019
196
1
TTF area
What if: $$(a – b)^2= a^2– b^2$$

Which values of a and b show that this is not true?

A. a = 1, b = 0 $\hspace{1cm}$ B. a = 1, b = 1

C. a = –1, b = –1 $\hspace{0.58cm}$ D. a = –1, b = 0

E. a = 1, b = –1

Thanks for those replies. How about this?
 
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