sqrt 16a^4x^2
- Thread starter Matholdie
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pka
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Take very special care to note the absolute value on x.
\(\displaystyle \L \sqrt {16a^4 x^2 } = 4a^2 \left| x \right|\)
We must use it because in the original problem what would be the value if a=-1 and x=-1? But what would be the value of the reduction without the absolute value?
Why do we not need then for the a?
\(\displaystyle \L \sqrt {16a^4 x^2 } = 4a^2 \left| x \right|\)
We must use it because in the original problem what would be the value if a=-1 and x=-1? But what would be the value of the reduction without the absolute value?
Why do we not need then for the a?