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(1-a)+3(1-a)^(2)=0 What are the solutions to the given equation? Choose 1 answer: (A) a=1 (B) a=(4)/(3) (c) a=1 and a=(4)/(3) (D) a=0 and a=-(1)/(3)

(1−a)+3(1−a)2=0 (1-a)+3(1-a)^{2}=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) a=1 a=1 \newline(B) a=43 a=\frac{4}{3} \newline(C) a=1 a=1 and a=43 a=\frac{4}{3} \newline(D) a=0 a=0 and a=−13 a=-\frac{1}{3}

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Q. (1−a)+3(1−a)2=0 (1-a)+3(1-a)^{2}=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) a=1 a=1 \newline(B) a=43 a=\frac{4}{3} \newline(C) a=1 a=1 and a=43 a=\frac{4}{3} \newline(D) a=0 a=0 and a=−13 a=-\frac{1}{3}
  1. Solving the quadratic equation: Now, we need to solve the quadratic equation 3a2−7a+4=03a^2 - 7a + 4 = 0. We can do this by factoring, completing the square, or using the quadratic formula. Let's try factoring first.\newline(3a−4)(a−1)=0(3a - 4)(a - 1) = 0\newlineSet each factor equal to zero and solve for aa.\newline3a−4=03a - 4 = 0 or a−1=0a - 1 = 0
  2. Factoring the quadratic equation: Solve the first equation 3a−4=03a - 4 = 0 for aa.3a=43a = 4a=43a = \frac{4}{3}
  3. Solving the first equation: Solve the second equation a−1=0a - 1 = 0 for aa.\newlinea=1a = 1
  4. Solving the second equation: We have found two solutions to the equation: a=43a = \frac{4}{3} and a=1a = 1. These are the values of aa that satisfy the original equation.

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