Question

Решить уравнение (х^2)-6х) ^2) -80=11((х^2)-6х)

Answer 1

(x^2-6x)^2 -80=11(x^2-6x)

(x^2-6X)^2-80-11(x^2-6x)=0

(x^2-6x)^2-11(x^2-6x)-80=0

x^2-6x= 16

x^2-6x= -5

x=8

x= -2

x=5

x=1

Вроде так.

Answer 2

To solve the quadratic equation $(x^2 - 6x)^2 - 80 = 11(x^2 - 6x)$, let's first make it easier to work with by introducing a substitution. Let $y = (x^2 - 6x)$. Now the equation becomes $y^2 - 80 = 11y$.

Now, let's rewrite it as a quadratic equation in terms of $y$:

$y^2 - 11y - 80 = 0$

To solve this quadratic equation for $y$, we can use factoring or the quadratic formula. In this case, factoring is simpler:

$y^2 - 11y - 80 = (y - 16)(y + 5) = 0$

Now, set each factor equal to zero and solve for $y$:

1. $y - 16 = 0 \Rightarrow y = 16$
2. $y + 5 = 0 \Rightarrow y = -5$

Now that we have two possible values for $y$, let's substitute them back into the equation $y = (x^2 - 6x)$:

1. If $y = 16$:

$16 = x^2 - 6x$

To solve this, we can set it to zero and factor it:

$x^2 - 6x - 16 = 0$

Now, use the quadratic formula to find the solutions for $x$:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1}$ $x = \frac{6 \pm \sqrt{36 + 64}}{2}$ $x = \frac{6 \pm \sqrt{100}}{2}$ $x = \frac{6 \pm 10}{2}$

This gives us two possible values for $x$:

a) $x = \frac{6 + 10}{2} = 8$ b) $x = \frac{6 - 10}{2} = -2$

2. If $y = -5$:

$-5 = x^2 - 6x$

Again, set it to zero and factor:

$x^2 - 6x + 5 = 0$

Using the quadratic formula:

$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$ $x = \frac{6 \pm \sqrt{36 - 20}}{2}$ $x = \frac{6 \pm \sqrt{16}}{2}$ $x = \frac{6 \pm 4}{2}$

This gives us two more possible values for $x$:

c) $x = \frac{6 + 4}{2} = 5$ d) $x = \frac{6 - 4}{2} = 1$

So, the solutions to the original equation are $x = 8$, $x = -2$, $x = 5$, and $x = 1$.