Let $a=x^2+4$ . How do you rewrite $(x^2+4)^2 +32 = 12x^2+48$ in terms of $a$ and set it equal to zero.?

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Answer 1 · 2017-02-22T06:08:17

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

$(x^2+4)^2+32=12x^2+48$ is equivalent to (factorizing RHS)

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

But as $a=x^2+4$ , this can be written as

$a^2+32=12xxa$

or $a^2+32=12a$

or $a^2-12a+32=0$

Let a=x^2+4a=x^2+4. How do you rewrite (x^2+4)^2 +32 = 12x^2+48(x^2+4)^2 +32 = 12x^2+48 in terms of aa and set it equal to zero.?