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

Expand squared term: First, let's expand the squared term (3x-1)². (3x-1)² = (3x-1)(3x-1) = 9x² - 3x - 3x + 1 = 9x² - 6x + 1Substitute expanded term: Now, substitute the expanded term back into the original equation. 9x² - 6x + 1 + 12x - 4 = 0Combine like terms: Combine like terms. 9x² + 6x - 3 = 0Solve using quadratic formula: This is a quadratic equation, and we can solve for x by factoring, completing the square, or using the quadratic formula. The equation does not factor easily, so let's use the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a), where a = 9, b = 6, and c = -3.Calculate discriminant: First, calculate the discriminant (b² - 4ac). Discriminant = 6² - 4(9)(-3) = 36 + 108 = 144Apply quadratic formula: Since the discriminant is positive, there are two real solutions. Now, apply the quadratic formula. x = (-6 ± √144) / (2 * 9) x = (-6 ± 12) / 18Calculate possible values: Calculate the two possible values for x. x₁ = (-6 + 12) / 18 = 6 / 18 = 1/3 x₂ = (-6 - 12) / 18 = -18 / 18 = -1

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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: $\newline$ $(3x-1)^{2}+12x-4=0$ $\newline$ $x=$

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Algebra 1

Solve linear equations: mixed review

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Q. Find one value of

x

that is a solution to the equation:

\newline

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

\newline

x=

Expand squared term: First, let's expand the squared term $(3x-1)^2$ . $(3x-1)^2 = (3x-1)(3x-1) = 9x^2 - 3x - 3x + 1 = 9x^2 - 6x + 1$
Substitute expanded term: Now, substitute the expanded term back into the original equation. $\newline$ $9x^2 - 6x + 1 + 12x - 4 = 0$
Combine like terms: Combine like terms. $\newline$ $9x^2 + 6x - 3 = 0$
Solve using quadratic formula: This is a quadratic equation, and we can solve for $x$ by factoring, completing the square, or using the quadratic formula. The equation does not factor easily, so let's use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 9$ , $b = 6$ , and $c = -3$ .
Calculate discriminant: First, calculate the discriminant $b^2 - 4ac$ . $\newline$ Discriminant = $6^2 - 4(9)(-3) = 36 + 108 = 144$
Apply quadratic formula: Since the discriminant is positive, there are two real solutions. Now, apply the quadratic formula. $\newline$ $x = \frac{-6 \pm \sqrt{144}}{2 \times 9}$ $\newline$ $x = \frac{-6 \pm 12}{18}$
Calculate possible values: Calculate the two possible values for $x$ . $x_1 = \frac{-6 + 12}{18} = \frac{6}{18} = \frac{1}{3}$ $x_2 = \frac{-6 - 12}{18} = \frac{-18}{18} = -1$