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Let a = x2 + 4. Use a to find the solutions for the following equation:

(x2 + 4)2 + 32 = 12x2 + 48
Which of the following are solutions for x?​

User AndHeiberg
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1 Answer

6 votes

The solutions for x are -2, 0, 2

Solution:

Given that,


\text { Let } a = x^2 + 4

Given equation is:


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


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


\text { Subtsitute } a = x^2 + 4 \text{ in above equation }


a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0


\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }


a^2 -4a - 8a + 32 = 0

Taking "a" as common term from first two terms and taking "-8" as common from last two terms


a(a-4)-8(a - 4) = 0


\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0


\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8


\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4


\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2


\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0


\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2

User VikramV
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