Quadratic Equation Question: How does 4 become -4 when v^2 = 4v + 96 becomes v^2 - 4v - 96 = 0?

Redneck Aussie

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Hey lads

Can someone please explain to me how the 4 in the below equation becomes a negative 4 in this step?
Thank you

11554
 

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The same way the "96" became "-96", subtraction!

You have \(\displaystyle v^2= 4v+96\). To get "0" on the right side, subtract "4v+ 96" from both sides:
\(\displaystyle v^2- (4v+96)= 4v+ 96- (4v+ 96)\)
\(\displaystyle v^2- 4v- 96= 0\)
 
If two quantities are equal, like v^2 and 4v+96, when you subtract them (v^2 -(4v+96) = v^2 - 4v - 96) then you get 0 (v^2 - 4v - 96 = 0)
 
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