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echo9000
Joined: 10 Feb 2023
Last visit: 09 Apr 2024
Posts: 15
Own Kudos:
Given Kudos: 9
Schools: Sloan '25 (WL)
19 Feb 2023, 06:19
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Hey guys,
\((x − 25)^2 = x^2 − 10x + 25\)
\((x + 5)^2*(x − 5)^2 = (x − 5)^2\)
Why can't we divide both sides by \((x − 5)^2\)?
\((x − 25)^2 = x^2 − 10x + 25\)
\((x + 5)^2*(x − 5)^2 = (x − 5)^2\)
Why can't we divide both sides by \((x − 5)^2\)?
19 Feb 2023, 08:53
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echo9000
You cannot reduce \((x + 5)^2*(x − 5)^2 = (x − 5)^2\) by \((x − 5)^2\) because \((x − 5)^2\) can be 0 and we cannot divide by 0. By doing so you loose a root, namely \((x − 5)^2=0\) --> x = 5.
Never reduce equation by a variable (or expression with a variable), if you are not certain that the variable (or expression with the variable) doesn't equal to zero. We cannot divide by zero.
07 Mar 2023, 05:42
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parkhydel
\((x^2 – 25)^2 = x^2 – 10x + 25\)
\([(x+5)(x-5)]^2= (x-5)^2\)
\((x+5)^2(x-5)^2= (x-5)^2\)
One solution is x=5, which will make both sides equal to 0.
If x≠5, then (x-5) is nonzero, allowing us to safely divide both sides by \((x-5)^2\):
\(\frac{(x+5)^2(x-5)^2}{(x-5)^2}= \frac{(x-5)^2}{(x-5)^2}\)
\((x+5)^2 = 1\)
The resulting equation is valid if x+5=1 or x+5=-1, yielding two additional values for x and bringing the total number of solutions to 3.
