Question

(x^2 –16)^2 + (x2 + x – 4)^2=0

Answer 1

(x² - 16)² + (x² + x - 4)² = 0   ⇔  

{ x² - 16 =0 ;         { x = ± 4 ;

{ x² + x - 4  = 0 .   { x ₁ , ₂  = (-1 ± √17)/2 .           x ∈  ∅

Ответ :  уравнение не имеет решения

P.S.  X²  ≥  0 для всех действительных   X .

Answer 2

(x² –16)^2 + (x² +x-4)²=0

Разложим на  множители

(х²-16)²=((х-4)*(х+4))²=(х-4)²*(х+4)²

x² + x-4=0 ; х=(-1±√17)/2

Сумма квадратов двух выражений равна нулю. когда оба выражения равны нулю, но если первое выражение равно нулю. тогда х=±4  но при х=±4 второе не обращается в нуль , значит, нет корней.

Answer 3

The equation you provided is:

$(x^2 - 16)^2 + (x^2 + x - 4)^2 = 0$

Let's solve it step by step.

First, let's look at the terms in the parentheses.

$x^2 - 16$ can be factored as $(x + 4)(x - 4)$.

$x^2 + x - 4$ cannot be factored any further.

Now, let's rewrite the equation:

$(x + 4)^2 (x - 4)^2 + (x^2 + x - 4)^2 = 0$

Next, let's expand these terms:

$(x + 4)^2 (x - 4)^2 = (x + 4)(x + 4)(x - 4)(x - 4)$

$= (x^2 + 8x + 16)(x^2 - 8x + 16)$

$= x^4 + 8x^3 + 16x^2 - 8x^3 - 64x^2 - 128x + 16x^2 + 128x + 256$

$= x^4 - 32x^2 + 256$

Next, we have $(x^2 + x - 4)^2$, which can be expanded as:

$(x^2 + x - 4)^2 = (x^2 + x - 4)(x^2 + x - 4)$

$= x^4 + 2x^3 - 3x^2 - 8x + 16$

Now, let's rewrite the original equation:

$x^4 - 32x^2 + 256 + x^4 + 2x^3 - 3x^2 - 8x + 16 = 0$

$2x^4 + 2x^3 - 35x^2 - 8x + 272 = 0$

This is a quartic (4th degree) polynomial equation. Solving it can be quite complex, and it may not have real roots. If you're looking for numerical solutions, you would typically use numerical methods or a computer algebra system to find the roots.

Keep in mind that in some cases, polynomial equations like this may have complex roots, meaning they involve imaginary numbers.