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

$$8 x ^ { 2 } - 8 x - 1 = 0$$

x uchun yechish

$$x=\frac{\sqrt{6}}{4}+\frac{1}{2}\approx 1,112372436$$

$$x=-\frac{\sqrt{6}}{4}+\frac{1}{2}\approx -0,112372436$$

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.

$$8x^{2}-8x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 8 ni a, -8 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 8\left(-1\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-32\left(-1\right)}}{2\times 8}$$

-32 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64+32}}{2\times 8}$$

64 ni 32 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{96}}{2\times 8}$$

96 ning kvadrat ildizini chiqarish.

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

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±4\sqrt{6}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{8±4\sqrt{6}}{16}$$

x=\frac{8±4\sqrt{6}}{16} tenglamasini yeching, bunda ± musbat. 8 ni 4\sqrt{6} ga qo'shish.

$$x=\frac{4\sqrt{6}+8}{16}$$

8+4\sqrt{6} ni 16 ga bo'lish.

$$x=\frac{\sqrt{6}}{4}+\frac{1}{2}$$

x=\frac{8±4\sqrt{6}}{16} tenglamasini yeching, bunda ± manfiy. 8 dan 4\sqrt{6} ni ayirish.

$$x=\frac{8-4\sqrt{6}}{16}$$

8-4\sqrt{6} ni 16 ga bo'lish.

$$x=-\frac{\sqrt{6}}{4}+\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{6}}{4}+\frac{1}{2} x=-\frac{\sqrt{6}}{4}+\frac{1}{2}$$

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.

$$8x^{2}-8x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$8x^{2}-8x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$8x^{2}-8x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$8x^{2}-8x=1$$

Ikki tarafini 8 ga bo‘ling.

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

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

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

-8 ni 8 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

$$x^{2}-x+\frac{1}{4}=\frac{1}{8}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{8} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-x+\frac{1}{4}=\frac{3}{8}$$

x^{2}-x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{6}}{4} x-\frac{1}{2}=-\frac{\sqrt{6}}{4}$$

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

$$x=\frac{\sqrt{6}}{4}+\frac{1}{2} x=-\frac{\sqrt{6}}{4}+\frac{1}{2}$$