(x-2)^2geq7 eching

$$( x - 2 ) ^ { 2 } \geq 7$$

x uchun yechish

$$x\in (-\infty,2-\sqrt{7}]\cup [\sqrt{7}+2,\infty)$$

Yechim qadamlari

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.

$$\left(x-2\right)^{2}=0$$

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 -4 ni va c uchun -3 ni ayiring.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{4±2\sqrt{7}}{2}$$

x=\frac{4±2\sqrt{7}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{7}+2\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\left(\sqrt{7}+2\right) va x-\left(2-\sqrt{7}\right) ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\left(\sqrt{7}+2\right) va x-\left(2-\sqrt{7}\right) ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\left(\sqrt{7}+2\right)\leq 0 x-\left(2-\sqrt{7}\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq 2-\sqrt{7}.

$$x\leq 2-\sqrt{7}$$

x-\left(\sqrt{7}+2\right) va x-\left(2-\sqrt{7}\right) ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x-\left(2-\sqrt{7}\right)\geq 0 x-\left(\sqrt{7}+2\right)\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \sqrt{7}+2.

$$x\geq \sqrt{7}+2$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq 2-\sqrt{7}\text{; }x\geq \sqrt{7}+2$$