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Levart [38]
3 years ago
12

Find the non-extraneous solutions of the square root of the quantity x plus 6 minus 5 equals quantity x plus 1.

Mathematics
1 answer:
denis23 [38]3 years ago
3 0
(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous

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Evaluate y = 2x + 1 when x = 1
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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
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Step-by-step explanation:

<u>Step 1: Define</u>

y = 2x + 1

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<u>Step 2: Evaluate</u>

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3 years ago
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Answer:  The graph is attached.

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y=\dfrac{1}{x},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)\\\\\\y=\dfrac{5}{x}+6.~~~~~~~~~~~~~~~~~~~~~~~~~(B)

In the attached figure, the graphs of both (A) and (B) are shown. We can easily see see from there, the shapes of both the graphs are same.

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Thus, the comparison can be seen in the figure very clearly.

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3 years ago
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