6x^2-8x=0 eching

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

x uchun yechish

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1,333333333$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(6x-8\right)=0$$

Tenglamani yechish uchun x=0 va 6x-8=0 ni yeching.

$$x=0 x=\frac{4}{3}$$

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.

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

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

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

\left(-8\right)^{2} ning kvadrat ildizini chiqarish.

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

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±8}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{8±8}{12}$$

x=\frac{8±8}{12} tenglamasini yeching, bunda ± musbat. 8 ni 8 ga qo'shish.

$$x=\frac{16}{12}$$

\frac{16}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{4}{3}$$

x=\frac{8±8}{12} tenglamasini yeching, bunda ± manfiy. 8 dan 8 ni ayirish.

$$x=\frac{0}{12}$$

0 ni 12 ga bo'lish.

$$x=0$$

Tenglama yechildi.

$$x=\frac{4}{3} x=0$$

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.

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

Ikki tarafini 6 ga bo‘ling.

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

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{8}{6}\right)x=\frac{0}{6}$$

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

$$x^{2}-\frac{4}{3}x=\frac{0}{6}$$

0 ni 6 ga bo'lish.

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

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

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

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

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

x^{2}-\frac{4}{3}x+\frac{4}{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.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-\frac{2}{3}=\frac{2}{3} x-\frac{2}{3}=-\frac{2}{3}$$

\frac{2}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{4}{3} x=0$$