x^2-6/8x-1/2=0 eching

$${ x }^{ 2 } - \frac{ 6 }{ 8 } x- \frac{ 1 }{ 2 } =0$$

x uchun yechish

$$x = \frac{\sqrt{41} + 3}{8} \approx 1,17539053$$

$$x=\frac{3-\sqrt{41}}{8}\approx -0,42539053$$

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.

$$x^{2}-\frac{3}{4}x-\frac{1}{2}=0$$

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{4} kvadratini chiqarish.

$$x=\frac{-\left(-\frac{3}{4}\right)±\sqrt{\frac{9}{16}-4\left(-\frac{1}{2}\right)}}{2}$$

-4 ni -\frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-\left(-\frac{3}{4}\right)±\sqrt{\frac{9}{16}+2}}{2}$$

\frac{9}{16} ni 2 ga qo'shish.

$$x=\frac{-\left(-\frac{3}{4}\right)±\sqrt{\frac{41}{16}}}{2}$$

\frac{41}{16} ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-\frac{3}{4}\right)±\frac{\sqrt{41}}{4}}{2}$$

-\frac{3}{4} ning teskarisi \frac{3}{4} ga teng.

$$x=\frac{\frac{3}{4}±\frac{\sqrt{41}}{4}}{2}$$

x=\frac{\frac{3}{4}±\frac{\sqrt{41}}{4}}{2} tenglamasini yeching, bunda ± musbat. \frac{3}{4} ni \frac{\sqrt{41}}{4} ga qo'shish.

$$x=\frac{\sqrt{41}+3}{2\times 4}$$

\frac{3+\sqrt{41}}{4} ni 2 ga bo'lish.

$$x=\frac{\sqrt{41}+3}{8}$$

x=\frac{\frac{3}{4}±\frac{\sqrt{41}}{4}}{2} tenglamasini yeching, bunda ± manfiy. \frac{3}{4} dan \frac{\sqrt{41}}{4} ni ayirish.

$$x=\frac{3-\sqrt{41}}{2\times 4}$$

\frac{3-\sqrt{41}}{4} ni 2 ga bo'lish.

$$x=\frac{3-\sqrt{41}}{8}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}+3}{8} x=\frac{3-\sqrt{41}}{8}$$

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.

$$x^{2}-\frac{3}{4}x-\frac{1}{2}=0$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -\frac{1}{2} ayirilsa 0 qoladi.

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

0 dan -\frac{1}{2} ni ayirish.

$$x^{2}-\frac{3}{4}x=\frac{1}{2}$$

-\frac{3}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{8} olish uchun. Keyin, -\frac{3}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{8} kvadratini chiqarish.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{1}{2}+\frac{9}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{9}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{41}{64}$$

x^{2}-\frac{3}{4}x+\frac{9}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{41}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Qisqartirish.

$$x-\frac{3}{8}=\frac{\sqrt{41}}{8} x-\frac{3}{8}=-\frac{\sqrt{41}}{8}$$

\frac{3}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{41}+3}{8} x=\frac{3-\sqrt{41}}{8}$$