Question

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?​

Answer 1

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`