Question

(x^2-64)^2+(x^2+5x-24)^2=0

Answer 1

$(x^{2} -64)^{2} +(x^{2} +5x-24)^{2}=0$

$a^{2} +b^{2} =0$, если a=0 и b=0

$1) x^{2} -64=0\\x^{2} =64\\x_{1} =8; x_{2} =-8$

$2) x^{2} +5x-24=0\\D=25+24*4=121\\x_{1}= \frac{-5+11}{2}=3\\x_{2} =\frac{-5-11}{2} =-8$

Ответ: -8; 3; 8

Answer 2

The given equation is a polynomial equation involving two terms raised to the power of 2. You want to find the solutions for x that satisfy this equation:

(x^2 - 64)^2 + (x^2 + 5x - 24)^2 = 0

To solve this equation, we can start by analyzing each term separately and then finding the values of x that make each term equal to zero.

Let's start with the first term: (x^2 - 64)^2 = 0

This term will only be equal to zero if its square, (x^2 - 64), is equal to zero. So we have: x^2 - 64 = 0

Adding 64 to both sides: x^2 = 64

Taking the square root of both sides: x = ±√64

x = ±8

Now, let's move on to the second term: (x^2 + 5x - 24)^2 = 0

Similarly, this term will be equal to zero only if its square, (x^2 + 5x - 24), is equal to zero. So we have: x^2 + 5x - 24 = 0

Factoring the quadratic equation: (x + 8)(x - 3) = 0

This gives us two possible solutions: x + 8 = 0 => x = -8 x - 3 = 0 => x = 3

So, the solutions for the given equation are x = -8, x = 8, and x = 3.

However, it's important to double-check the original equation to make sure these solutions are valid. Substituting each value of x back into the original equation should confirm whether they are indeed solutions.