12x+15x^2=0 eching

$$12 x + 15 x ^ { 2 } = 0$$

x uchun yechish

$$x=-\frac{4}{5}=-0,8$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(12+15x\right)=0$$

Tenglamani yechish uchun x=0 va 12+15x=0 ni yeching.

$$x=0 x=-\frac{4}{5}$$

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.

$$15x^{2}+12x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 15 ni a, 12 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}}}{2\times 15}$$

12^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-12±12}{2\times 15}$$

2 ni 15 marotabaga ko'paytirish.

$$x=\frac{-12±12}{30}$$

x=\frac{-12±12}{30} tenglamasini yeching, bunda ± musbat. -12 ni 12 ga qo'shish.

$$x=\frac{0}{30}$$

0 ni 30 ga bo'lish.

$$x=0$$

x=\frac{-12±12}{30} tenglamasini yeching, bunda ± manfiy. -12 dan 12 ni ayirish.

$$x=-\frac{24}{30}$$

\frac{-24}{30} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{4}{5}$$

Tenglama yechildi.

$$x=0 x=-\frac{4}{5}$$

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.

$$15x^{2}+12x=0$$

Ikki tarafini 15 ga bo‘ling.

$$\frac{15x^{2}+12x}{15}=\frac{0}{15}$$

15 ga bo'lish 15 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{12}{15}x=\frac{0}{15}$$

\frac{12}{15} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{4}{5}x=\frac{0}{15}$$

0 ni 15 ga bo'lish.

$$x^{2}+\frac{4}{5}x=0$$

\frac{4}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{5} olish uchun. Keyin, \frac{2}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\left(\frac{2}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{5} kvadratini chiqarish.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

x^{2}+\frac{4}{5}x+\frac{4}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Qisqartirish.

$$x+\frac{2}{5}=\frac{2}{5} x+\frac{2}{5}=-\frac{2}{5}$$

Tenglamaning ikkala tarafidan \frac{2}{5} ni ayirish.

$$x=0 x=-\frac{4}{5}$$