Question

(х^2+х-3)^2-12(х^2+х-3)+24=0

Answer 1

Ответ:

(x^2 + x - 3)^2 - 12(x^2 + x - 3) + 24 = 0

Решим это уравнение методом замены переменной (х).

Пусть: (x^2 + x - 3) = t. Тогда (x^2 + x - 3)^2 = t^2.

t^2 - 12t + 24 = 0

D = 144 - 96 = 48

√D = √48 = 4√3

t1 = (12 - 4√3) / 2 = 6 - 2√3

t2 = (12 + 4√3) / 2 = 6 + 2√3

Вернёмся к начальному уравнению и замене:

1) x^2 + x - 3 = 6 - 2√3

x^2 + x - 9 + 2√3 = 0

x1,2 = - 1/2 +- √(0,5)^2 - (-9 + 2√3) = -1/2 +- √1/4 + 9 - 2√3 = -1/2 +- √37/4 - 2√3

2) x^2 + x - 3 = 6 + 2√3

x^2 + x - 9 - 2√3 = 0

x1,2 = -1/2 +- √(1/4) - (-9 - 2√3) = -1/2 +- √37/4 + 2√3

Ответ:

$x_{1} = -\frac{1}{2} + \sqrt{\frac{37}{4} -2\sqrt{3} }$

$x_{2} = -\frac{1}{2} - \sqrt{\frac{37}{4} -2\sqrt{3} }$

Answer 2

To solve the equation $(x^2 + x - 3)^2 - 12(x^2 + x - 3) + 24 = 0$, you can use substitution to simplify it. Let $y = x^2 + x - 3$. Then, the equation becomes:

$y^2 - 12y + 24 = 0$

Now, we have a quadratic equation in terms of $y$, which is easier to solve. You can use the quadratic formula to find the solutions:

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

In this case, $a = 1$, $b = -12$, and (c = 24. Plug these values into the formula:

$y = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(24)}}{2(1)}$

Simplify this:

$y = \frac{12 \pm \sqrt{144 - 96}}{2}$ $y = \frac{12 \pm \sqrt{48}}{2}$

Now, simplify the square root:

$y = \frac{12 \pm 4\sqrt{3}}{2}$

Divide both terms by 2:

$y = 6 \pm 2\sqrt{3}$

Now that we've found the possible values of $y$, we can substitute them back into the original equation to find the corresponding values of $x$:

1. If $y = 6 + 2\sqrt{3}$, then:

$x^2 + x - 3 = 6 + 2\sqrt{3}$

Now, solve this quadratic equation for $x$.

2. If $y = 6 - 2\sqrt{3}$, then:

$x^2 + x - 3 = 6 - 2\sqrt{3}$

Solve this quadratic equation for $x$.

You'll have two sets of solutions for $x$, corresponding to the two possible values of $y$.

Answer 3

To solve the quadratic equation $(x^2 + x - 3)^2 - 12(x^2 + x - 3) + 24 = 0$, we can first simplify it by noticing that $(x^2 + x - 3)$ is a common factor in all the terms. Let's denote $(x^2 + x - 3)$ as $y$:

So, $y = x^2 + x - 3$.

The equation becomes:

$y^2 - 12y + 24 = 0$

Now, you can solve this quadratic equation for $y$. We can use the quadratic formula:

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

In this case, $a = 1$, $b = -12$, and $c = 24$.

Plugging these values into the quadratic formula:

$y = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(24)}}{2(1)}$

Simplify:

$y = \frac{12 \pm \sqrt{144 - 96}}{2}$ $y = \frac{12 \pm \sqrt{48}}{2}$ $y = \frac{12 \pm 4\sqrt{3}}{2}$

Now, we have two potential values for $y$:

1. $y = 6 + 2\sqrt{3}$
2. $y = 6 - 2\sqrt{3}$

However, remember that $y = x^2 + x - 3$. So, we can set up two separate equations for each value of $y$:

1. $x^2 + x - 3 = 6 + 2\sqrt{3}$
2. $x^2 + x - 3 = 6 - 2\sqrt{3}$

Let's solve these equations separately:

1. $x^2 + x - 3 = 6 + 2\sqrt{3}$

Add $3$ to both sides:

$x^2 + x = 9 + 2\sqrt{3}$

Now, we can solve for $x$:

$x^2 + x - 9 - 2\sqrt{3} = 0$

You can use the quadratic formula again to find the values of $x$:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-9 - 2\sqrt{3})}}{2(1)}$

Simplify:

$x = \frac{-1 \pm \sqrt{1 + 36 + 8\sqrt{3}}}{2}$