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3 years ago

You are supposed to distribute to only one term (the first).

Mathematics

2 answers:

Tom [10]3 years ago

6 0

Answer:

true

Step-by-step explanation:

There is no distribution (unless you count substitution) in the second set of terms.

Edit: I think I get what your question is now. You are supposed to distribute to every term inside the parenthesis (false)

algol [13]3 years ago

4 0

True, follow PEMDAS

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We will proceed to solve each case to determine the solution of the problem.

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case a) is not the solution of the problem

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Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square. Remember to balance the equation by adding the same constants to each side.

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Rewrite as perfect squares

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therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

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$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

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Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

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Rewrite as perfect squares

$(x-1)^{2}=5$

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case d) is the solution of the problem

therefore

the answer is

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