Find one value of x that is a solution to the equation: (3x+7)^(2)=-6x-14 x=

Write Equation: Write down the given equation. (3x + 7)² = -6x - 14Expand Equation: Expand the left side of the equation. (3x + 7)(3x + 7) = -6x - 14 9x² + 21x + 21x + 49 = -6x - 14 Combine like terms. 9x² + 42x + 49 = -6x - 14Combine Like Terms: Move all terms to one side to set the equation to zero. 9x² + 42x + 49 + 6x + 14 = 0 Combine like terms. 9x² + 48x + 63 = 0Move Terms to One Side: Factor the quadratic equation. We need to find two numbers that multiply to (9 * 63) and add up to 48. The numbers 9 and 63 already satisfy this condition. (3x + 9)(3x + 7) = 0Factor Quadratic Equation: Solve for x by setting each factor equal to zero. 3x + 9 = 0 or 3x + 7 = 0 For the first factor: 3x = -9 x = -3 For the second factor: 3x = -7 x = -7/3Solve for x: Check the solutions in the original equation. For x = -3: (3(-3) + 7)² = -6(-3) - 14 (-9 + 7)² = 18 - 14 (-2)² = 4 4 = 4 (True) For x = -7/3: (3(-7/3) + 7)² = -6(-7/3) - 14 (-7 + 7)² = 14 - 14 0² = 0 0 = 0 (True) Both solutions are valid.

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Find one value of
x that is a solution to the equation:

(3x+7)^(2)=-6x-14

x=

Find one value of $x$ that is a solution to the equation: $\newline$ $(3x+7)^{2}=-6x-14$ $\newline$ $x=$

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Math Problems

Algebra 1

Solve a quadratic equation by factoring

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Q. Find one value of

x

that is a solution to the equation:

\newline

(3x+7)^{2}=-6x-14

\newline

x=

Write Equation: Write down the given equation. $\newline$ $(3x + 7)^2 = -6x - 14$
Expand Equation: Expand the left side of the equation. $\newline$ $(3x + 7)(3x + 7) = -6x - 14$ $\newline$ $9x^2 + 21x + 21x + 49 = -6x - 14$ $\newline$ Combine like terms. $\newline$ $9x^2 + 42x + 49 = -6x - 14$
Combine Like Terms: Move all terms to one side to set the equation to zero. $\newline$ $9x^2 + 42x + 49 + 6x + 14 = 0$ $\newline$ Combine like terms. $\newline$ $9x^2 + 48x + 63 = 0$
Move Terms to One Side: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $(9 \times 63)$ and add up to $48$ . $\newline$ The numbers $9$ and $63$ already satisfy this condition. $\newline$ $(3x + 9)(3x + 7) = 0$
Factor Quadratic Equation: Solve for x by setting each factor equal to zero. $\newline$ $3x + 9 = 0$ or $3x + 7 = 0$ $\newline$ For the first factor: $\newline$ $3x = -9$ $\newline$ $x = -3$ $\newline$ For the second factor: $\newline$ $3x = -7$ $\newline$ $x = -\frac{7}{3}$
Solve for x: Check the solutions in the original equation. $\newline$ For $x = -3$ : $\newline$ $(3(-3) + 7)^2 = -6(-3) - 14$ $\newline$ $(-9 + 7)^2 = 18 - 14$ $\newline$ $(-2)^2 = 4$ $\newline$ $4 = 4$ (True) $\newline$ For $x = -\frac{7}{3}$ : $\newline$ $(3(-\frac{7}{3}) + 7)^2 = -6(-\frac{7}{3}) - 14$ $\newline$ $(-7 + 7)^2 = 14 - 14$ $\newline$ $0^2 = 0$ $\newline$ $0 = 0$ (True) $\newline$ Both solutions are valid.