t^2-6t+1geq0 eching

$$t ^ { 2 } - 6 t + 1 \geq 0$$

t uchun yechish

$$t\in (-\infty,3-2\sqrt{2}]\cup [2\sqrt{2}+3,\infty)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tengsizlikni yechish uchun chap tomon faktorini hisoblang. Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$t^{2}-6t+1=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun -6 ni va c uchun 1 ni ayiring.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 1}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{6±4\sqrt{2}}{2}$$

t=\frac{6±4\sqrt{2}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=2\sqrt{2}+3 t=3-2\sqrt{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(t-\left(2\sqrt{2}+3\right)\right)\left(t-\left(3-2\sqrt{2}\right)\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun t-\left(2\sqrt{2}+3\right) va t-\left(3-2\sqrt{2}\right) ikkalasi ≤0 yoki ≥0 boʻlishi kerak. t-\left(2\sqrt{2}+3\right) va t-\left(3-2\sqrt{2}\right) ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$t-\left(2\sqrt{2}+3\right)\leq 0 t-\left(3-2\sqrt{2}\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – t\leq 3-2\sqrt{2}.

$$t\leq 3-2\sqrt{2}$$

t-\left(2\sqrt{2}+3\right) va t-\left(3-2\sqrt{2}\right) ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$t-\left(3-2\sqrt{2}\right)\geq 0 t-\left(2\sqrt{2}+3\right)\geq 0$$

Ikkala tengsizlikning mos yechimi – t\geq 2\sqrt{2}+3.

$$t\geq 2\sqrt{2}+3$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$t\leq 3-2\sqrt{2}\text{; }t\geq 2\sqrt{2}+3$$