Solve the quadratic by factoring. x^(2)+8x-17=7x-5 Answer: x=

Write Standard Form: Write the quadratic equation in standard form. To solve the quadratic by factoring, we need to have it in the form ax^2 + bx + c = 0. We start by moving all terms to one side of the equation. x^2 + 8x - 17 = 7x - 5 Subtract 7x from both sides: x^2 + 8x - 7x - 17 + 5 = 0 x^2 + x - 12 = 0Factor Quadratic Equation: Factor the quadratic equation. We need to find two numbers that multiply to -12 and add up to 1 (the coefficient of x). The numbers that satisfy these conditions are 4 and -3 because: 4 * (-3) = -12 4 + (-3) = 1 So we can factor the quadratic as: (x + 4)(x - 3) = 0Solve for x: Solve for x. To find the solutions to the equation, we set each factor equal to zero and solve for x. First factor: x + 4 = 0 x = -4 Second factor: x - 3 = 0 x = 3

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Algebra 1

Factor quadratics with leading coefficient 1

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Q. Solve the quadratic by factoring.

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x^{2}+8 x-17=7 x-5

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Answer:

x=

Write Standard Form: Write the quadratic equation in standard form. $\newline$ To solve the quadratic by factoring, we need to have it in the form $ax^2 + bx + c = 0$ . We start by moving all terms to one side of the equation. $\newline$ $x^2 + 8x - 17 = 7x - 5$ $\newline$ Subtract $7x$ from both sides: $\newline$ $x^2 + 8x - 7x - 17 + 5 = 0$ $\newline$ $x^2 + x - 12 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $-12$ and add up to $1$ (the coefficient of $x$ ). $\newline$ The numbers that satisfy these conditions are $4$ and $-3$ because: $\newline$ $4 \times (-3) = -12$ $\newline$ $4 + (-3) = 1$ $\newline$ So we can factor the quadratic as: $\newline$ $(x + 4)(x - 3) = 0$
Solve for x: Solve for x. $\newline$ To find the solutions to the equation, we set each factor equal to zero and solve for $x$ . $\newline$ First factor: $\newline$ $x + 4 = 0$ $\newline$ $x = -4$ $\newline$ Second factor: $\newline$ $x - 3 = 0$ $\newline$ $x = 3$