Solve for \( b \) in the equation:
\[ 2b^2 + 4b - 2 = 0 \]
Answer (2)
[2b^2+4b-2=0\ 2(b^2+2b-1)=0\ 2(b^2+2b+1-2)=0\ 2((b+1)^2-2)=0\ 2(b+1)^2-4=0\ 2(b+1)^2=4\ (b+1)^2=2\ b+1=-\sqrt2 \vee b+1=\sqrt2\ b=-1-\sqrt2 \vee b=-1+\sqrt2
]
To solve the equation 2 b 2 + 4 b − 2 = 0 , first simplify it to b 2 + 2 b − 1 = 0 and then use the quadratic formula. The solutions are b = − 1 + 2 and b = − 1 − 2 .
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