x^2+2/3x-1/6=0 eching

$${ x }^{ 2 } + \frac{ 2 }{ 3 } x- \frac{ 1 }{ 6 } =0$$

x uchun yechish

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3}\approx 0,193712943$$

$$x=-\frac{\sqrt{10}}{6}-\frac{1}{3}\approx -0,86037961$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}+\frac{2}{3}x-\frac{1}{6}=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, \frac{2}{3} ni b va -\frac{1}{6} ni c bilan almashtiring.

$$x=\frac{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\left(-\frac{1}{6}\right)}}{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{3} kvadratini chiqarish.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\left(-\frac{1}{6}\right)}}{2}$$

-4 ni -\frac{1}{6} marotabaga ko'paytirish.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+\frac{2}{3}}}{2}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{4}{9} ni \frac{2}{3} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{10}{9}}}{2}$$

\frac{10}{9} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2}$$

x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2} tenglamasini yeching, bunda ± musbat. -\frac{2}{3} ni \frac{\sqrt{10}}{3} ga qo'shish.

$$x=\frac{\sqrt{10}-2}{2\times 3}$$

\frac{-2+\sqrt{10}}{3} ni 2 ga bo'lish.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3}$$

x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2} tenglamasini yeching, bunda ± manfiy. -\frac{2}{3} dan \frac{\sqrt{10}}{3} ni ayirish.

$$x=\frac{-\sqrt{10}-2}{2\times 3}$$

\frac{-2-\sqrt{10}}{3} ni 2 ga bo'lish.

$$x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3} x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$x^{2}+\frac{2}{3}x-\frac{1}{6}=0$$

\frac{1}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+\frac{2}{3}x-\frac{1}{6}-\left(-\frac{1}{6}\right)=-\left(-\frac{1}{6}\right)$$

O‘zidan -\frac{1}{6} ayirilsa 0 qoladi.

$$x^{2}+\frac{2}{3}x=-\left(-\frac{1}{6}\right)$$

0 dan -\frac{1}{6} ni ayirish.

$$x^{2}+\frac{2}{3}x=\frac{1}{6}$$

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{1}{6}+\left(\frac{1}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{6}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{6} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{18}$$

x^{2}+\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{5}{18}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{5}{18}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{\sqrt{10}}{6} x+\frac{1}{3}=-\frac{\sqrt{10}}{6}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3} x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$