8x^2-4x=18 eching

$$8 { x }^{ 2 } -4x=18$$

x uchun yechish

$$x = \frac{\sqrt{37} + 1}{4} \approx 1,770690633$$

$$x=\frac{1-\sqrt{37}}{4}\approx -1,270690633$$

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.

$$8x^{2}-4x=18$$

Tenglamaning ikkala tarafidan 18 ni ayirish.

$$8x^{2}-4x-18=18-18$$

O‘zidan 18 ayirilsa 0 qoladi.

$$8x^{2}-4x-18=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 8 ni a, -4 ni b va -18 ni c bilan almashtiring.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 8\left(-18\right)}}{2\times 8}$$

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 8\left(-18\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16-32\left(-18\right)}}{2\times 8}$$

-32 ni -18 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+576}}{2\times 8}$$

16 ni 576 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{592}}{2\times 8}$$

592 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±4\sqrt{37}}{2\times 8}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±4\sqrt{37}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{4±4\sqrt{37}}{16}$$

x=\frac{4±4\sqrt{37}}{16} tenglamasini yeching, bunda ± musbat. 4 ni 4\sqrt{37} ga qo'shish.

$$x=\frac{4\sqrt{37}+4}{16}$$

4+4\sqrt{37} ni 16 ga bo'lish.

$$x=\frac{\sqrt{37}+1}{4}$$

x=\frac{4±4\sqrt{37}}{16} tenglamasini yeching, bunda ± manfiy. 4 dan 4\sqrt{37} ni ayirish.

$$x=\frac{4-4\sqrt{37}}{16}$$

4-4\sqrt{37} ni 16 ga bo'lish.

$$x=\frac{1-\sqrt{37}}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{37}+1}{4} x=\frac{1-\sqrt{37}}{4}$$

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.

$$8x^{2}-4x=18$$

Ikki tarafini 8 ga bo‘ling.

$$\frac{8x^{2}-4x}{8}=\frac{18}{8}$$

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

$$x^{2}+\left(-\frac{4}{8}\right)x=\frac{18}{8}$$

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

$$x^{2}-\frac{1}{2}x=\frac{18}{8}$$

\frac{18}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{9}{4}+\left(-\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{4}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{9}{4} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{37}{16}$$

x^{2}-\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{37}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{37}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{\sqrt{37}}{4} x-\frac{1}{4}=-\frac{\sqrt{37}}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{37}+1}{4} x=\frac{1-\sqrt{37}}{4}$$