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galben [10]
3 years ago
15

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

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0

The solutions for x are -2, 0, 2

<em><u>Solution:</u></em>

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

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