The solutions for x are -2, 0, 2
Solution:
Given that,
[tex]\text { Let } a = x^2 + 4[/tex]
Given equation is:
[tex](x^2 + 4)^2 + 32 = 12x^2 + 48[/tex]
[tex](x^2 + 4)^2 + 32 = 12(x^2 + 4)[/tex]
[tex]\text { Subtsitute } a = x^2 + 4 \text{ in above equation }[/tex]
[tex]a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0[/tex]
[tex]\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }[/tex]
[tex]a^2 -4a - 8a + 32 = 0[/tex]
Taking "a" as common term from first two terms and taking "-8" as common from last two terms
[tex]a(a-4)-8(a - 4) = 0[/tex]
[tex]\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0[/tex]
[tex]\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8[/tex]
[tex]\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4[/tex]
[tex]\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2[/tex]
[tex]\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0[/tex]
[tex]\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2[/tex]
