8x^2-6x-4=0 eching

$$8 { x }^{ 2 } -6x-4=0$$

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.

$$8x^{2}-6x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 8 ni a, -6 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\left(-4\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\left(-4\right)}}{2\times 8}$$

-32 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+128}}{2\times 8}$$

36 ni 128 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{164}}{2\times 8}$$

164 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{41}}{2\times 8}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{41}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{6±2\sqrt{41}}{16}$$

x=\frac{6±2\sqrt{41}}{16} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{41} ga qo'shish.

$$x=\frac{2\sqrt{41}+6}{16}$$

6+2\sqrt{41} ni 16 ga bo'lish.

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

x=\frac{6±2\sqrt{41}}{16} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{41} ni ayirish.

$$x=\frac{6-2\sqrt{41}}{16}$$

6-2\sqrt{41} ni 16 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.

$$8x^{2}-6x-4=0$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$8x^{2}-6x-4-\left(-4\right)=-\left(-4\right)$$

O‘zidan -4 ayirilsa 0 qoladi.

$$8x^{2}-6x=-\left(-4\right)$$

0 dan -4 ni ayirish.

$$8x^{2}-6x=4$$

Ikki tarafini 8 ga bo‘ling.

$$\frac{8x^{2}-6x}{8}=\frac{4}{8}$$

8 ga bo'lish 8 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{6}{8}\right)x=\frac{4}{8}$$

\frac{-6}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{3}{4}x=\frac{4}{8}$$

\frac{4}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

-\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.

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