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$Find one value of x that is a solution to the equation: {:[(3x-1)^(2)+12 x-4=0],[x=◻]:}$

Find one value of $x$ that is a solution to the equation: $\newline$ $(3 x-1)^{2}+12 x-4=0$ $\newline$ $x=$

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Solve a quadratic equation by factoring

Full solution

Q. Find one value of

x

that is a solution to the equation:

\newline

(3 x-1)^{2}+12 x-4=0

\newline

x=

Write equation: Write down the given equation. $\newline$ $(3x-1)^2 + 12x - 4 = 0$
Expand squared term: Expand the squared term $(3x-1)^2$ .
$(3x-1)(3x-1) + 12x - 4 = 0$
$9x^2 - 3x - 3x + 1 + 12x - 4 = 0$
$9x^2 + 6x - 3 = 0$
Factor quadratic equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $(9)(-3) = -27$ and add up to $6$ . The numbers that satisfy these conditions are $9$ and $-3$ . $\newline$ $9x^2 + 9x - 3x - 3 = 0$
Group and factor by grouping: Group the terms and factor by grouping. $\newline$ $(9x^2 + 9x) - (3x + 3) = 0$ $\newline$ $9x(x + 1) - 3(x + 1) = 0$
Factor out common binomial: Factor out the common binomial factor $(x + 1)$ . $\newline$ $(9x - 3)(x + 1) = 0$
Solve for x: Set each factor equal to zero and solve for x. $\newline$ $9x - 3 = 0$ or $x + 1 = 0$ $\newline$ For $9x - 3 = 0$ : $\newline$ $9x = 3$ $\newline$ $x = \frac{3}{9}$ $\newline$ $x = \frac{1}{3}$ $\newline$ For $x + 1 = 0$ : $\newline$ $x = -1$
Write solutions: Write down the solutions. $\newline$ $x = \frac{1}{3}$ or $x = -1$ $\newline$ We can choose either value as a solution to the original equation.

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