9n^2+21n=0 eching

$$9 n ^ { 2 } + 21 n = 0$$

n uchun yechish

$$n = -\frac{7}{3} = -2\frac{1}{3} \approx -2,333333333$$

$$n=0$$

Faktorlash qadamlari

n omili.

$$n\left(9n+21\right)=0$$

Tenglamani yechish uchun n=0 va 9n+21=0 ni yeching.

$$n=0 n=-\frac{7}{3}$$

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.

$$9n^{2}+21n=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, 21 ni b va 0 ni c bilan almashtiring.

$$n=\frac{-21±\sqrt{21^{2}}}{2\times 9}$$

21^{2} ning kvadrat ildizini chiqarish.

$$n=\frac{-21±21}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$n=\frac{-21±21}{18}$$

n=\frac{-21±21}{18} tenglamasini yeching, bunda ± musbat. -21 ni 21 ga qo'shish.

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

0 ni 18 ga bo'lish.

$$n=0$$

n=\frac{-21±21}{18} tenglamasini yeching, bunda ± manfiy. -21 dan 21 ni ayirish.

$$n=-\frac{42}{18}$$

\frac{-42}{18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=-\frac{7}{3}$$

Tenglama yechildi.

$$n=0 n=-\frac{7}{3}$$

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.

$$9n^{2}+21n=0$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9n^{2}+21n}{9}=\frac{0}{9}$$

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

$$n^{2}+\frac{21}{9}n=\frac{0}{9}$$

\frac{21}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n^{2}+\frac{7}{3}n=\frac{0}{9}$$

0 ni 9 ga bo'lish.

$$n^{2}+\frac{7}{3}n=0$$

\frac{7}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{6} olish uchun. Keyin, \frac{7}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}+\frac{7}{3}n+\left(\frac{7}{6}\right)^{2}=\left(\frac{7}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{6} kvadratini chiqarish.

$$n^{2}+\frac{7}{3}n+\frac{49}{36}=\frac{49}{36}$$

n^{2}+\frac{7}{3}n+\frac{49}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(n+\frac{7}{6}\right)^{2}=\frac{49}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{7}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Qisqartirish.

$$n+\frac{7}{6}=\frac{7}{6} n+\frac{7}{6}=-\frac{7}{6}$$

Tenglamaning ikkala tarafidan \frac{7}{6} ni ayirish.

$$n=0 n=-\frac{7}{3}$$