Solve the quadratic by factoring. x^(2)-4x-41=-9x+9 Answer: x=

Move to Standard Form: Write the equation in standard form by moving all terms to one side. x^2 - 4x - 41 = -9x + 9 Add 9x to both sides and subtract 9 from both sides to get: x^2 + 5x - 50 = 0Factor the Equation: Factor the quadratic equation. We need to find two numbers that multiply to -50 and add up to 5. The numbers 10 and -5 satisfy these conditions because: 10 * -5 = -50 10 + (-5) = 5 So we can write the factored form as: (x + 10)(x - 5) = 0Solve for x: Solve for x by setting each factor equal to zero. x + 10 = 0 or x - 5 = 0 Solving each equation gives us: x = -10 or x = 5

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Math Problems

Algebra 1

Factor quadratics with leading coefficient 1

Full solution

Q. Solve the quadratic by factoring.

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x^{2}-4 x-41=-9 x+9

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Answer:

x=

Move to Standard Form: Write the equation in standard form by moving all terms to one side. $\newline$ $x^2 - 4x - 41 = -9x + 9$ $\newline$ Add $9x$ to both sides and subtract $9$ from both sides to get: $\newline$ $x^2 + 5x - 50 = 0$
Factor the Equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $-50$ and add up to $5$ . $\newline$ The numbers $10$ and $-5$ satisfy these conditions because: $\newline$ $10 \times -5 = -50$ $\newline$ $10 + (-5) = 5$ $\newline$ So we can write the factored form as: $\newline$ $(x + 10)(x - 5) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 10 = 0$ or $x - 5 = 0$ $\newline$ Solving each equation gives us: $\newline$ $x = -10$ or $x = 5$