Question

X4+5x2-8=0 dau 10puncte

Answer 1

Ответ:

Пошаговое объяснение:

x^4+5x^2-24=0 Замена:x^2=t t^2+5t-24=0 D=b^2-4ac=5^2-4*1*(-24)=25+96=121 t1=-5+11/2=3 t2=-5-11=-8 t1=x^2=3 x=±√3

Answer 2

To solve the quadratic equation $X^4 + 5X^2 - 8 = 0$, we can use a substitution to make it a quadratic equation in terms of a new variable. Let's use the substitution $Y = X^2$. Then the equation becomes:

$Y^2 + 5Y - 8 = 0$

Now, we can solve this quadratic equation for Y using the quadratic formula:

The quadratic formula is given by: $Y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where, for our equation $Y^2 + 5Y - 8 = 0$, the coefficients are: $a = 1$, $b = 5$, $c = -8$.

Now, let's calculate the values of Y:

$Y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}$

$Y = \frac{-5 \pm \sqrt{25 + 32}}{2}$

$Y = \frac{-5 \pm \sqrt{57}}{2}$

So, we have two potential solutions for Y:

1. $Y = \frac{-5 + \sqrt{57}}{2}$
2. $Y = \frac{-5 - \sqrt{57}}{2}$

Now, remember that we used the substitution $Y = X^2$. So, to find the values of X, we need to take the square root of each of the Y solutions:

1. $X^2 = \frac{-5 + \sqrt{57}}{2}$ $X = \pm \sqrt{\frac{-5 + \sqrt{57}}{2}}$

2. $X^2 = \frac{-5 - \sqrt{57}}{2}$ $X = \pm \sqrt{\frac{-5 - \sqrt{57}}{2}}$

Now, we have four potential solutions for X, accounting for both positive and negative square roots. However, we need to check whether these solutions are valid by substituting them back into the original equation $X^4 + 5X^2 - 8 = 0$ and verifying if they satisfy it.

Please note that the solutions might involve complex numbers, as the expression under the square root ($57$) is not a perfect square. Without numerical values, it's not possible to provide the exact solutions, but the process outlined above should give you the steps to solve the equation.