Let a=x^2+4. Use a to find the solutions for the following equation: (x^2+4)^2+32=12x^2+48. Which one of the following are solutions for x? Select any/all that apply. -8, -2, 4, 0, 2, -4, 8

Answers

Answer 1

Answer: $(x^2+4)^2+32=12x^2+48 \n(x^2+4)^2+32=12(x^2+4) \ \ \ |-12(x^2+4) \n(x^2+4)^2-12(x^2+4)+32=0 \n\hbox{substitute a for } x^2+4: \na^2-12a+32=0 \na^2-4a-8a+32=0 \na(a-4)-8(a-4)=0 \n(a-8)(a-4)=0 \na-8=0 \ \lor \ a-4=0 \na=8 \ \lor \ a=4 \n \n\hbox{substitute 8 and 4 for a and solve for x:} \na=8 \n\Downarrow \n8=x^2+4 \ \ \ |-4 \n4=x^2 \nx=-2 \ \lor \ x=2 \n \na=4 \n\Downarrow \n4=x^2+4 \ \ \ |-4 \n0=x^2 \nx=0 \n \n\boxed{x=-2 \hbox{ or } x=0 \hbox{ or } x=2}$

The solutions for x are -2, 0, 2.

Answer 2

Answer:

Answer:

-2,0,2

Step-by-step explanation:

The given equation is:

$(x^2+4)^(2)+32=12x^2+48$

⇒ $(x^2+4)^(2)+32=12(x^2+4)$

Substituting $(x^2+4)=a$ in the above equation, we get

⇒ $a^(2)+32=12a$

⇒ $a^2-12a+32=0$

⇒ $a^2-4a-8a+32=0$

⇒ $a(a-4)-8(a-4)=0$

⇒ $(a-8)(a-4)=0$

⇒ $a=8,4$

Now, $(x^2+4)=a$ , then substituting the value of a in this equation,

$x^(2)+4=8$ and $x^2+4=4$

⇒ $x^(2)+4=8$

⇒ $x={\pm}2$ and

⇒ $x^(2)+4=4$

⇒ $x=0$

Thus, the value of x are -2,0 and 2.

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Answers

As a result of the sale, two chairs can be purchased at the discounted price of $92.16.

After the store bought the chairs for $36, they gave it a markup of 60%. The new price is:

= Cost of chair + Markup

= 36 + (60% x 36)

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Find out more at brainly.com/question/18410539.

starts at 36....marks up by 60%
36 + 0.60(36) = 36 + 21.60 = 57.60
now they mark it down by 20%
57.60 - 0.20(57.60) = 57.60 - 11.52 = 46.08

46.08 * 2 = 92.16

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

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