Question

(х^2+х) ^2-6(х^2+х) +8=0​

Answer 1

$\bf (x^2+x) ^2-6(x^2+x) +8=0$

$t {}^{2} - 6t + 8 = 0$

$\fbox{ t = 2 \\ t = 4}$

$\fbox{x {}^{2} + x = 2 \\ x {}^{2} + x = 4}$

$\fbox{x = - 2 \\ x = 1 \\ x = \dfrac{ - 1 + \sqrt{17} }{2} \\ x = \dfrac{ - 1 - \sqrt{17} }{2} }$

Ответ:

$\fbox{x_{1} = - 2 \\ x_{2} = 1 \\ x_{3} = \dfrac{ - 1 + \sqrt{17} }{2} \\ x_{4} = \dfrac{ - 1 - \sqrt{17} }{2} }$

Answer 2

Ответ:

Пошаговое объяснение:

x²+x=t

t²-6t+8=0

t1=4; t2=2

1)x²+x=4⇒x²+x-4=0⇒D=1+16=17⇒x1=(-1+√17)/2; x2=(-1-√17)/2

2)x²+x=2⇒x²+x-2=0⇒x3=1; x4=-2

Answer 3

To solve the quadratic equation $(x^2 + x)^2 - 6(x^2 + x) + 8 = 0$, you can use substitution to simplify it and then solve for (x.

Let $y = x^2 + x$. Now, we can rewrite the equation as:

$y^2 - 6y + 8 = 0$

This is a quadratic equation in $y$, which is easier to solve. To factor or solve for (y, we can rewrite the equation as:

$(y - 2)(y - 4) = 0$

Now, we can solve for (y\ by setting each factor equal to zero:

1. $y - 2 = 0$ $y = 2$

2. $y - 4 = 0$ $y = 4$

Now that we have found the possible values of y\, we can revert to the substitution \(y = x^2 + x:

For $y = 2$:

$x^2 + x = 2$

This is a quadratic equation in (x, and we can solve it using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1\, b = 1\, and c = -2$. Plugging these values into the formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}$

$x = \frac{-1 \pm \sqrt{1 + 8}}{2}$

$x = \frac{-1 \pm \sqrt{9}}{2}$

Now, take the square root:

$x = \frac{-1 \pm 3}{2}$

This gives two solutions:

1. $x = \frac{-1 + 3}{2} = 1$
2. $x = \frac{-1 - 3}{2} = -2$

For $y = 4$:

$x^2 + x = 4$

Again, we can solve this as a quadratic equation:

$x = \frac{-1 \pm \sqrt{1 + 4(4)}}{2}$

$x = \frac{-1 \pm \sqrt{17}}{2}$

So, the solutions to the original equation are $x = 1$ and $x = -2$ for $y = 2$, and $x = \frac{-1 + \sqrt{17}}{2}$ and $x = \frac{-1 - \sqrt{17}}{2}$ for $y = 4$.