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elena55 [62]
3 years ago
7

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

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

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

<u>case d)</u> 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.

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

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

Rewrite as perfect squares

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

therefore

case d) is the solution of the problem

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<u>the answer is</u>

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