[tex](x-1)^{2}[/tex][tex]\leq[/tex]9

Предыдущий
вопрос Следующий
вопрос

To solve the inequality tex^{2} \leq 9[/tex], we first need to expand tex^{2}[/tex].

tex^{2} = x^{2} - 2x + 1[/tex]

Now, we have [tex]x^{2} - 2x + 1 \leq 9[/tex].

Subtracting 9 from both sides, we get:

[tex]x^{2} - 2x - 8 \leq 0[/tex]

Now, we need to find the roots of the quadratic equation [tex]x^{2} - 2x - 8 = 0[/tex].

The roots can be found using the quadratic formula:

[tex]x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 41(-8)}}{2*1}[/tex]
[tex]x = \frac{2 \pm \sqrt{4 + 32}}{2}[/tex]
[tex]x = \frac{2 \pm \sqrt{36}}{2}[/tex]
[tex]x = \frac{2 \pm 6}{2}[/tex]

So, the roots are [tex]x = -2[/tex] and [tex]x = 4[/tex].

The inequality [tex]x^{2} - 2x - 8 \leq 0[/tex] can now be written in interval notation as:

[tex][-2, 4][/tex]

Therefore, the solution to the inequality tex^{2} \leq 9[/tex] is [tex]-2 \leq x \leq 4[/tex].