Question

PLEASE HELP!!20 pts question

Answer 1

value of x is x=2, x=-2 and x=0

Step-by-step explanation:

Let a = x^2+4. Use a to find the solution for the following equation:

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

Solving

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

Taking 12 common from LHS

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

Let a=x^2+4

$(a)^2+32=12(a)\\a^2+32=12a\\a^2-12a+32=0$

Factoring the equation

$a^2-8a-4a+32=0\\a(a-8)-4(a-8)=0\\(a-8)(a-4)=0\\a-8=0\,\,and\,\,a-4=0\\a=8\,\,and\,\,a=4$

Putting value of a i.e x^2+4

$x^2+4=8\,\,and\,\,x^2+4=4\\x^2=8-4\,\,and\,\,x^2=4-4\\x^2=4\,\,and\,\,x^2=0\\Taking\,\,square\,\,root\,\,at:\\x=\pm 2\,\,and\,\,x=0$

So, value of x is x=2, x=-2 and x=0

Keywords: Solving Equations

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