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2 votes
I don't understand how it gets the result
28b_(2) and
b_(2)^(2) in the fourth equation.

I don't understand how it gets the result 28b_(2) and b_(2)^(2) in the fourth equation-example-1
I don't understand how it gets the result 28b_(2) and b_(2)^(2) in the fourth equation-example-1
I don't understand how it gets the result 28b_(2) and b_(2)^(2) in the fourth equation-example-2

1 Answer

3 votes

Answer + Step-by-step explanation:

your question is about equation 4 :


b_(1)=14-b_(2)

Then


\left( b_(1)\right)^(2) =\left( 14-b_(2)\right)^(2) \ \text{(this is a special product )}

Then


\left( b_(1)\right)^(2) =14^(2)-2* 14 * b_(2)+ (b_2)^2

Then


\left( b_(1)\right)^(2) =196-28 b_(2)+ (b_2)^2

User Wizart
by
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