n^2+3n-12=0 eching

$$n ^ { 2 } + 3 n - 12 = 0$$

n uchun yechish

$$n = \frac{\sqrt{57} - 3}{2} \approx 2,274917218$$

$$n=\frac{-\sqrt{57}-3}{2}\approx -5,274917218$$

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.

$$n^{2}+3n-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 3 ni b va -12 ni c bilan almashtiring.

$$n=\frac{-3±\sqrt{3^{2}-4\left(-12\right)}}{2}$$

3 kvadratini chiqarish.

$$n=\frac{-3±\sqrt{9-4\left(-12\right)}}{2}$$

-4 ni -12 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{9+48}}{2}$$

9 ni 48 ga qo'shish.

$$n=\frac{-3±\sqrt{57}}{2}$$

n=\frac{-3±\sqrt{57}}{2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{57} ga qo'shish.

$$n=\frac{\sqrt{57}-3}{2}$$

n=\frac{-3±\sqrt{57}}{2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{57} ni ayirish.

$$n=\frac{-\sqrt{57}-3}{2}$$

Tenglama yechildi.

$$n=\frac{\sqrt{57}-3}{2} n=\frac{-\sqrt{57}-3}{2}$$

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.

$$n^{2}+3n-12=0$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$n^{2}+3n-12-\left(-12\right)=-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$n^{2}+3n=-\left(-12\right)$$

0 dan -12 ni ayirish.

$$n^{2}+3n=12$$

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

$$n^{2}+3n+\left(\frac{3}{2}\right)^{2}=12+\left(\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$n^{2}+3n+\frac{9}{4}=12+\frac{9}{4}$$

12 ni \frac{9}{4} ga qo'shish.

$$n^{2}+3n+\frac{9}{4}=\frac{57}{4}$$

n^{2}+3n+\frac{9}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(n+\frac{3}{2}\right)^{2}=\frac{57}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{3}{2}\right)^{2}}=\sqrt{\frac{57}{4}}$$

Qisqartirish.

$$n+\frac{3}{2}=\frac{\sqrt{57}}{2} n+\frac{3}{2}=-\frac{\sqrt{57}}{2}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$n=\frac{\sqrt{57}-3}{2} n=\frac{-\sqrt{57}-3}{2}$$