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1-9b^(2)=0 What are the solutions to the given equation? Choose 1 answer: (A) b=-(1)/(3) and b=(1)/(3) (B) b=-(1)/(9) and b=(1)/(9) (C) b=(1)/(3) (D) b=(1)/(9)

1−9b2=0 1-9 b^{2}=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) b=−13 b=-\frac{1}{3} and b=13 b=\frac{1}{3} \newline(B) b=−19 b=-\frac{1}{9} and b=19 b=\frac{1}{9} \newline(C) b=13 b=\frac{1}{3} \newline(D) b=19 b=\frac{1}{9}

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Q. 1−9b2=0 1-9 b^{2}=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) b=−13 b=-\frac{1}{3} and b=13 b=\frac{1}{3} \newline(B) b=−19 b=-\frac{1}{9} and b=19 b=\frac{1}{9} \newline(C) b=13 b=\frac{1}{3} \newline(D) b=19 b=\frac{1}{9}
  1. Isolate term with variable b: First, let's isolate the term containing the variable b.\newline1−9b2=01 - 9b^2 = 0\newlineAdd 9b29b^2 to both sides to get:\newline1=9b21 = 9b^2
  2. Divide both sides by 99: Next, we divide both sides by 99 to solve for b2b^2.\newline19=b2\frac{1}{9} = b^2
  3. Take square root of both sides: Now, we take the square root of both sides to solve for b b . Remember that taking the square root of a number gives us two solutions, one positive and one negative.b=±19 b = \pm\sqrt{\frac{1}{9}}
  4. Simplify square root of 19\frac{1}{9}: Simplify the square root of 19\frac{1}{9}.b=±13b = \pm\frac{1}{3}
  5. Find two solutions for b: We have found two solutions for b, which are b = \frac{11}{33} and b = -\frac{11}{33}.

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