Solve for all values of x : (x-5)^(2)+(x-5)=0 Answer: x=

Set up equation: Set up the equation to solve for x. We are given the equation (x-5)² + (x-5) = 0. We need to find all values of x that satisfy this equation.Factor the equation: Factor the equation. Notice that (x-5) is a common factor in both terms of the equation. We can factor it out: (x-5)((x-5) + 1) = 0 This simplifies to: (x-5)(x-5+1) = 0 (x-5)(x-4) = 0Apply zero product property: Apply the zero product property. If the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x: x-5 = 0 or x-4 = 0Solve for x: Solve each equation for x. For the first equation: x-5 = 0 x = 5 For the second equation: x-4 = 0 x = 4Check solutions: Check the solutions in the original equation. Substitute x = 5 into the original equation: (5-5)² + (5-5) = 0 0² + 0 = 0 0 = 0 This is true, so x = 5 is a solution. Substitute x = 4 into the original equation: (4-5)² + (4-5) = 0 (-1)² + (-1) = 0 1 - 1 = 0 0 = 0 This is true, so x = 4 is also a solution.

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Solve for all values of
x :

(x-5)^(2)+(x-5)=0
Answer:
x=

Solve for all values of $x$ : $\newline$ $(x-5)^{2}+(x-5)=0$ $\newline$ Answer: $x=$

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Algebra 2

Composition of linear and quadratic functions: find a value

Full solution

Q. Solve for all values of

x

\newline

(x-5)^{2}+(x-5)=0

\newline

Answer:

x=

Set up equation: Set up the equation to solve for $x$ . We are given the equation $(x-5)^2 + (x-5) = 0$ . We need to find all values of $x$ that satisfy this equation.
Factor the equation: Factor the equation. $\newline$ Notice that $(x-5)$ is a common factor in both terms of the equation. We can factor it out: $\newline$ $(x-5)((x-5) + 1) = 0$ $\newline$ This simplifies to: $\newline$ $(x-5)(x-5+1) = 0$ $\newline$ $(x-5)(x-4) = 0$
Apply zero product property: Apply the zero product property. $\newline$ If the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for $x$ : $\newline$ $x-5 = 0$ or $x-4 = 0$
Solve for x: Solve each equation for x. $\newline$ For the first equation: $\newline$ $x-5 = 0$ $\newline$ $x = 5$ $\newline$ For the second equation: $\newline$ $x-4 = 0$ $\newline$ $x = 4$
Check solutions: Check the solutions in the original equation. $\newline$ Substitute $x = 5$ into the original equation: $\newline$ $(5-5)^2 + (5-5) = 0$ $\newline$ $0^2 + 0 = 0$ $\newline$ $0 = 0$ $\newline$ This is true, so $x = 5$ is a solution. $\newline$ Substitute $x = 4$ into the original equation: $\newline$ $(4-5)^2 + (4-5) = 0$ $\newline$ $(-1)^2 + (-1) = 0$ $\newline$ $1 - 1 = 0$ $\newline$ $0 = 0$ $\newline$ This is true, so $x = 4$ is also a solution.