2b^2-4b+1>0 eching

$$2 b ^ { 2 } - 4 b + 1 > 0$$

b uchun yechish

$$b\in \left(-\infty,-\frac{\sqrt{2}}{2}+1\right)\cup \left(\frac{\sqrt{2}}{2}+1,\infty\right)$$

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.

$$2b^{2}-4b+1=0$$

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

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

Hisoblarni amalga oshiring.

$$b=\frac{4±2\sqrt{2}}{4}$$

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

$$b=\frac{\sqrt{2}}{2}+1 b=-\frac{\sqrt{2}}{2}+1$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$2\left(b-\left(\frac{\sqrt{2}}{2}+1\right)\right)\left(b-\left(-\frac{\sqrt{2}}{2}+1\right)\right)>0$$

Koʻpaytma musbat boʻlishi uchun b-\left(\frac{\sqrt{2}}{2}+1\right) va b-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi yo manfiy, yo musbat boʻlishi kerak. b-\left(\frac{\sqrt{2}}{2}+1\right) va b-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi manfiy boʻlganda, yechimini toping.

$$b-\left(\frac{\sqrt{2}}{2}+1\right)<0 b-\left(-\frac{\sqrt{2}}{2}+1\right)<0$$

Ikkala tengsizlikning mos yechimi – b<-\frac{\sqrt{2}}{2}+1.

$$b<-\frac{\sqrt{2}}{2}+1$$

b-\left(\frac{\sqrt{2}}{2}+1\right) va b-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi musbat boʻlganda, yechimini toping.

$$b-\left(-\frac{\sqrt{2}}{2}+1\right)>0 b-\left(\frac{\sqrt{2}}{2}+1\right)>0$$

Ikkala tengsizlikning mos yechimi – b>\frac{\sqrt{2}}{2}+1.

$$b>\frac{\sqrt{2}}{2}+1$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$b<-\frac{\sqrt{2}}{2}+1\text{; }b>\frac{\sqrt{2}}{2}+1$$