9x^2+4x=0 eching

$$9 x ^ { 2 } + 4 x = 0$$

x uchun yechish

$$x=-\frac{4}{9}\approx -0,444444444$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(9x+4\right)=0$$

Tenglamani yechish uchun x=0 va 9x+4=0 ni yeching.

$$x=0 x=-\frac{4}{9}$$

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.

$$9x^{2}+4x=0$$

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

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 9}$$

4^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-4±4}{18}$$

x=\frac{-4±4}{18} tenglamasini yeching, bunda ± musbat. -4 ni 4 ga qo'shish.

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

0 ni 18 ga bo'lish.

$$x=0$$

x=\frac{-4±4}{18} tenglamasini yeching, bunda ± manfiy. -4 dan 4 ni ayirish.

$$x=-\frac{8}{18}$$

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

$$x=-\frac{4}{9}$$

Tenglama yechildi.

$$x=0 x=-\frac{4}{9}$$

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.

$$9x^{2}+4x=0$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+4x}{9}=\frac{0}{9}$$

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

$$x^{2}+\frac{4}{9}x=\frac{0}{9}$$

0 ni 9 ga bo'lish.

$$x^{2}+\frac{4}{9}x=0$$

\frac{4}{9} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{9} olish uchun. Keyin, \frac{2}{9} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{9} kvadratini chiqarish.

$$x^{2}+\frac{4}{9}x+\frac{4}{81}=\frac{4}{81}$$

x^{2}+\frac{4}{9}x+\frac{4}{81} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{2}{9}\right)^{2}=\frac{4}{81}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{9}\right)^{2}}=\sqrt{\frac{4}{81}}$$

Qisqartirish.

$$x+\frac{2}{9}=\frac{2}{9} x+\frac{2}{9}=-\frac{2}{9}$$

Tenglamaning ikkala tarafidan \frac{2}{9} ni ayirish.

$$x=0 x=-\frac{4}{9}$$