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June 4th, 2019, 03:28 PM   #1
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Algebra what is the actual answer ?

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

Which values of a and b show that this is not true?
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June 4th, 2019, 03:59 PM   #2
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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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Last edited by Greens; June 4th, 2019 at 04:25 PM.
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June 4th, 2019, 04:15 PM   #3
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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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June 4th, 2019, 10:06 PM   #4
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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?

Last edited by skipjack; June 5th, 2019 at 08:04 AM.
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June 5th, 2019, 03:53 AM   #5
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June 5th, 2019, 05:28 AM   #6
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You seriously can't plug in the values and see whether it equals??
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