3n^2+47n-232=5 eching

$$3 n ^ { 2 } + 47 n - 232 = 5$$

n uchun yechish

$$n = \frac{\sqrt{5053} - 47}{6} \approx 4,014076135$$

$$n=\frac{-\sqrt{5053}-47}{6}\approx -19,680742802$$

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.

$$3n^{2}+47n-232=5$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$3n^{2}+47n-232-5=5-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$3n^{2}+47n-232-5=0$$

-232 dan 5 ni ayirish.

$$3n^{2}+47n-237=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 47 ni b va -237 ni c bilan almashtiring.

$$n=\frac{-47±\sqrt{47^{2}-4\times 3\left(-237\right)}}{2\times 3}$$

47 kvadratini chiqarish.

$$n=\frac{-47±\sqrt{2209-4\times 3\left(-237\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$n=\frac{-47±\sqrt{2209-12\left(-237\right)}}{2\times 3}$$

-12 ni -237 marotabaga ko'paytirish.

$$n=\frac{-47±\sqrt{2209+2844}}{2\times 3}$$

2209 ni 2844 ga qo'shish.

$$n=\frac{-47±\sqrt{5053}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$n=\frac{-47±\sqrt{5053}}{6}$$

n=\frac{-47±\sqrt{5053}}{6} tenglamasini yeching, bunda ± musbat. -47 ni \sqrt{5053} ga qo'shish.

$$n=\frac{\sqrt{5053}-47}{6}$$

n=\frac{-47±\sqrt{5053}}{6} tenglamasini yeching, bunda ± manfiy. -47 dan \sqrt{5053} ni ayirish.

$$n=\frac{-\sqrt{5053}-47}{6}$$

Tenglama yechildi.

$$n=\frac{\sqrt{5053}-47}{6} n=\frac{-\sqrt{5053}-47}{6}$$

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.

$$3n^{2}+47n-232=5$$

232 ni tenglamaning ikkala tarafiga qo'shish.

$$3n^{2}+47n-232-\left(-232\right)=5-\left(-232\right)$$

O‘zidan -232 ayirilsa 0 qoladi.

$$3n^{2}+47n=5-\left(-232\right)$$

5 dan -232 ni ayirish.

$$3n^{2}+47n=237$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3n^{2}+47n}{3}=\frac{237}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$n^{2}+\frac{47}{3}n=\frac{237}{3}$$

237 ni 3 ga bo'lish.

$$n^{2}+\frac{47}{3}n=79$$

\frac{47}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{47}{6} olish uchun. Keyin, \frac{47}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}+\frac{47}{3}n+\left(\frac{47}{6}\right)^{2}=79+\left(\frac{47}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{47}{6} kvadratini chiqarish.

$$n^{2}+\frac{47}{3}n+\frac{2209}{36}=79+\frac{2209}{36}$$

79 ni \frac{2209}{36} ga qo'shish.

$$n^{2}+\frac{47}{3}n+\frac{2209}{36}=\frac{5053}{36}$$

n^{2}+\frac{47}{3}n+\frac{2209}{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.

$$\left(n+\frac{47}{6}\right)^{2}=\frac{5053}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{47}{6}\right)^{2}}=\sqrt{\frac{5053}{36}}$$

Qisqartirish.

$$n+\frac{47}{6}=\frac{\sqrt{5053}}{6} n+\frac{47}{6}=-\frac{\sqrt{5053}}{6}$$

Tenglamaning ikkala tarafidan \frac{47}{6} ni ayirish.

$$n=\frac{\sqrt{5053}-47}{6} n=\frac{-\sqrt{5053}-47}{6}$$