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What do I need to do to get \[\frac{ c }{ a }-\frac{ b^2 }{ 4a^2 }\]=\[\frac{ 4ac-b^2 }{ 4a }\]
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u cant lol
these 2 expressions are not equal
I mean equal, I am trying to prove the quadratic formula by completing the square.
*dont
and my text say the first expression can be manipulated into the second.
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\[\frac{ c }{ a }-\frac{ b^2 }{ 4a^2 }=\frac{4ac-b^2}{4a^2}\]
ok how?
@telltoamit
c/a can be written as 4ac/4a^2
how did this get a medal, I am know the wiser.
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hehee lol
\[\frac{ c }{ a }-\frac{ b^2 }{ 4a^2 }\] 1. Multiply first fraction by 4a/4a to get: \[=\frac{4ac}{4a^2} -\frac{b^2}{4a^2}\] 2. Combine fractions to get: \[=\frac{4ac-b^2}{4a^2}\]3. You're finished
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