n^2+n-162=0 eching

$$n ^ { 2 } + n - 162 = 0$$

n uchun yechish

$$n = \frac{\sqrt{649} - 1}{2} \approx 12,237739203$$

$$n=\frac{-\sqrt{649}-1}{2}\approx -13,237739203$$

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.

$$n^{2}+n-162=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 1 ni b va -162 ni c bilan almashtiring.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-162\right)}}{2}$$

1 kvadratini chiqarish.

$$n=\frac{-1±\sqrt{1-4\left(-162\right)}}{2}$$

-4 ni -162 marotabaga ko'paytirish.

$$n=\frac{-1±\sqrt{1+648}}{2}$$

1 ni 648 ga qo'shish.

$$n=\frac{-1±\sqrt{649}}{2}$$

n=\frac{-1±\sqrt{649}}{2} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{649} ga qo'shish.

$$n=\frac{\sqrt{649}-1}{2}$$

n=\frac{-1±\sqrt{649}}{2} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{649} ni ayirish.

$$n=\frac{-\sqrt{649}-1}{2}$$

Tenglama yechildi.

$$n=\frac{\sqrt{649}-1}{2} n=\frac{-\sqrt{649}-1}{2}$$

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.

$$n^{2}+n-162=0$$

162 ni tenglamaning ikkala tarafiga qo'shish.

$$n^{2}+n-162-\left(-162\right)=-\left(-162\right)$$

O‘zidan -162 ayirilsa 0 qoladi.

$$n^{2}+n=-\left(-162\right)$$

0 dan -162 ni ayirish.

$$n^{2}+n=162$$

1 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{2} olish uchun. Keyin, \frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=162+\left(\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$n^{2}+n+\frac{1}{4}=162+\frac{1}{4}$$

162 ni \frac{1}{4} ga qo'shish.

$$n^{2}+n+\frac{1}{4}=\frac{649}{4}$$

n^{2}+n+\frac{1}{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.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{649}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{649}{4}}$$

Qisqartirish.

$$n+\frac{1}{2}=\frac{\sqrt{649}}{2} n+\frac{1}{2}=-\frac{\sqrt{649}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$n=\frac{\sqrt{649}-1}{2} n=\frac{-\sqrt{649}-1}{2}$$