(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)

Solving the quadratic equation: Now, we need to solve the quadratic equation 3a^2 - 7a + 4 = 0. We can do this by factoring, completing the square, or using the quadratic formula. Let's try factoring first. (3a - 4)(a - 1) = 0 Set each factor equal to zero and solve for a. 3a - 4 = 0 or a - 1 = 0Factoring the quadratic equation: Solve the first equation 3a - 4 = 0 for a. 3a = 4 a = 4/3Solving the first equation: Solve the second equation a - 1 = 0 for a. a = 1Solving the second equation: We have found two solutions to the equation: a = 4/3 and a = 1. These are the values of a that satisfy the original equation.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

(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)

$(1-a)+3(1-a)^{2}=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $a=1$ $\newline$ (B) $a=\frac{4}{3}$ $\newline$ (C) $a=1$ and $a=\frac{4}{3}$ $\newline$ (D) $a=0$ and $a=-\frac{1}{3}$

View step-by-step help

Home

Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

(1-a)+3(1-a)^{2}=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

a=1

\newline

(B)

a=\frac{4}{3}

\newline

(C)

a=1

and

a=\frac{4}{3}

\newline

(D)

a=0

and

a=-\frac{1}{3}

Solving the quadratic equation: Now, we need to solve the quadratic equation $3a^2 - 7a + 4 = 0$ . We can do this by factoring, completing the square, or using the quadratic formula. Let's try factoring first. $\newline$ $(3a - 4)(a - 1) = 0$ $\newline$ Set each factor equal to zero and solve for $a$ . $\newline$ $3a - 4 = 0$ or $a - 1 = 0$
Factoring the quadratic equation: Solve the first equation $3a - 4 = 0$ for $a$ . $3a = 4$ $a = \frac{4}{3}$
Solving the first equation: Solve the second equation $a - 1 = 0$ for $a$ . $\newline$ $a = 1$
Solving the second equation: We have found two solutions to the equation: $a = \frac{4}{3}$ and $a = 1$ . These are the values of $a$ that satisfy the original equation.