# Find x for (x^2 - 16)^2 + 2x^2 – 28 = 0

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We'll substitute x^2 by a and we'll transform the expression into a quadratic equation.

(x^2 - 16)^2 + 2x^2 – 28 = 0

x^2 = a

(a-16)^2 + 2a - 28 = 0

We'll expand the square:

a^2 - 32a + 256 + 2a - 28 = 0

We'll combine like terms:

a^2 - 30a + 228 = 0

We'll spply the quadratic formula:

a1 = [30+sqrt(900 - 912)]/2

a1 = (30 + 2isqrt3)/2

a1 = 15 + isqrt3

a2 = 15 - isqrt3

But x^2 = a

x1 = +sqrt(15 + isqrt3)

x2 = -sqrt(15 + isqrt3)

x3 = +sqrt(15 - isqrt3)

x4 = -sqrt(15 - isqrt3)

But (15 + isqrt3) and (15 - isqrt3) are complex numbers, so there are no real solutions for the equation (x^2 - 16)^2 + 2x^2 – 28 = 0

We have to solve (x^2 - 16)^2 + 2x^2 – 28 = 0

=> (x^2 - 16)^2 + 2x^2 – 32 + 4 = 0

let Y = x^2 – 16

=> Y^2 + 2 Y + 4 =0

=> (Y + 2)^2 = 4

=> Y + 2= 2 or Y + 2 =-2

For Y+ 2 =2

Y =0

=> x^2 – 16 =0

=> x^2 = 16

=> x = 4 or x = -4

For Y+2 = -2

Y = -4

=> x^2 -16 =-4

=> x^2 = 12

=> x = 2 sqrt 3 or x = -2 sqrt 3

**Therefore we have four values of x: 2, -2, 2 sqrt 3 and -2 sqrt 3**

Find x for (x^2 - 16)^2 + 2x^2 – 28 = 0

Put x^2 -16 = t.Then the equation becomes:

Then 2x^2 = 2(t+16).

So the given equation becomes:

t^2+2(t+16)-28 = 0

t^2+2t +4 = 0

t = {-2+or-sqrt(2^2-4*1*4)}/2 , Or

t1 = {-1 +sqrt (-3)}. Or t2 = {-1-sqrt(-3)}

Therefore ,

x^2-16 = -1 +sqrt(-3) gives x = 16-1 +sqrt(-3(

x^2 = 15+sqrt(-3) , or x^2 = 15-sqrt(-3).

Therefore there are 4 complex roots which are below:

x1 = sqrt{15+sqrt(-3)} or x2= sqr{15-sqrt(-3)}

x3 = -sqrt{15-sqrt(-3)} or x4 = -sqrt{15-sqrt(-3)}