x(x-a)-b(a+b)=0 In the given equation, a and b are constants. What are the solutions to the equation? Choose 1 answer: (A) x=-b and x=(a-b) (B) x=-b and x=(a+b) (c) x=a and x=(a-b) (D) x=a and x=(a+b)

Expand and Rewrite Equation: First, let's expand the equation to see if it can be factored easily. x(x-a) - b(a+b) = 0 x^2 - ax - ab - b^2 = 0Factor Quadratic Equation: Now, we need to factor the quadratic equation if possible. We are looking for two numbers that multiply to give -ab - b^2 and add up to -a. However, we notice that the equation is not in a standard quadratic form, and it seems that we cannot factor it directly. Instead, let's look at the original equation and try to factor by grouping. x(x-a) - b(a+b) = 0 (x - b)(x - a) = 0Apply Zero Product Property: We have factored the equation into two binomials. Now we can use the zero product property, which states that if a product of two factors is zero, then at least one of the factors must be zero. So we set each factor equal to zero: x - b = 0 or x - a = 0Solve for x: Solving each equation for x gives us the possible solutions: x - b = 0 => x = b x - a = 0 => x = aCheck Answer Choices: We have found two potential solutions for x. However, we need to check the answer choices to see which one matches our solutions. (A) x=-b and x=(a-b) (B) x=-b and x=(a+b) (C) x=a and x=(a-b) (D) x=a and x=(a+b)Identify Mistake: By comparing our solutions x = b and x = a with the answer choices, we see that none of the answer choices exactly match our solutions. There seems to be a mistake in our factoring step. Let's go back and correct it.

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x(x-a)-b(a+b)=0
In the given equation,
a and
b are constants. What are the solutions to the equation?
Choose 1 answer:
(A)
x=-b and
x=(a-b)
(B)
x=-b and
x=(a+b)
(c)
x=a and
x=(a-b)
(D)
x=a and
x=(a+b)

$x(x-a)-b(a+b)=0$ $\newline$ In the given equation, $a$ and $b$ are constants. What are the solutions to the equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-b$ and $x=(a-b)$ $\newline$ (B) $x=-b$ and $x=(a+b)$ $\newline$ (C) $x=a$ and $x=(a-b)$ $\newline$ (D) $x=a$ and $x=(a+b)$

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Algebra 2

Write equations of cosine functions using properties

Full solution

x(x-a)-b(a+b)=0

\newline

In the given equation,

a

and

b

are constants. What are the solutions to the equation?

\newline

Choose

1

answer:

\newline

(A)

x=-b

and

x=(a-b)

\newline

(B)

x=-b

and

x=(a+b)

\newline

(C)

x=a

and

x=(a-b)

\newline

(D)

x=a

and

x=(a+b)

Expand and Rewrite Equation: First, let's expand the equation to see if it can be factored easily. $\newline$ $x(x-a) - b(a+b) = 0$ $\newline$ $x^2 - ax - ab - b^2 = 0$
Factor Quadratic Equation: Now, we need to factor the quadratic equation if possible. We are looking for two numbers that multiply to give $-ab - b^2$ and add up to $-a$ . However, we notice that the equation is not in a standard quadratic form, and it seems that we cannot factor it directly. Instead, let's look at the original equation and try to factor by grouping. $\newline$ $x(x-a) - b(a+b) = 0$ $\newline$ $(x - b)(x - a) = 0$
Apply Zero Product Property: We have factored the equation into two binomials. Now we can use the zero product property, which states that if a product of two factors is zero, then at least one of the factors must be zero. $\newline$ So we set each factor equal to zero: $\newline$ $x - b = 0$ or $x - a = 0$
Solve for x: Solving each equation for x gives us the possible solutions: $\newline$ $x - b = 0 \Rightarrow x = b$ $\newline$ $x - a = 0 \Rightarrow x = a$
Check Answer Choices: We have found two potential solutions for $x$ . However, we need to check the answer choices to see which one matches our solutions. $\newline$ (A) $x=-b$ and $x=(a-b)$ $\newline$ (B) $x=-b$ and $x=(a+b)$ $\newline$ (C) $x=a$ and $x=(a-b)$ $\newline$ (D) $x=a$ and $x=(a+b)$
Identify Mistake: By comparing our solutions $x = b$ and $x = a$ with the answer choices, we see that none of the answer choices exactly match our solutions. There seems to be a mistake in our factoring step. Let's go back and correct it.