9x^2+6x=0 eching

$$9 x ^ { 2 } + 6 x = 0$$

x uchun yechish

$$x=-\frac{2}{3}\approx -0,666666667$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(9x+6\right)=0$$

Tenglamani yechish uchun x=0 va 9x+6=0 ni yeching.

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

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.

$$9x^{2}+6x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, 6 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 9}$$

6^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±6}{18}$$

x=\frac{-6±6}{18} tenglamasini yeching, bunda ± musbat. -6 ni 6 ga qo'shish.

$$x=\frac{0}{18}$$

0 ni 18 ga bo'lish.

$$x=0$$

x=\frac{-6±6}{18} tenglamasini yeching, bunda ± manfiy. -6 dan 6 ni ayirish.

$$x=-\frac{12}{18}$$

\frac{-12}{18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

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

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.

$$9x^{2}+6x=0$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+6x}{9}=\frac{0}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{9}x=\frac{0}{9}$$

\frac{6}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{2}{3}x=\frac{0}{9}$$

0 ni 9 ga bo'lish.

$$x^{2}+\frac{2}{3}x=0$$

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

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

x^{2}+\frac{2}{3}x+\frac{1}{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.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{1}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{1}{3} x+\frac{1}{3}=-\frac{1}{3}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

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