x^2+33x=6 eching

$$x ^ { 2 } + 33 x = 6$$

x uchun yechish

$$x=\frac{\sqrt{1113}-33}{2}\approx 0,180827318$$

$$x=\frac{-\sqrt{1113}-33}{2}\approx -33,180827318$$

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.

$$x^{2}+33x=6$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$x^{2}+33x-6=6-6$$

O‘zidan 6 ayirilsa 0 qoladi.

$$x^{2}+33x-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 33 ni b va -6 ni c bilan almashtiring.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-6\right)}}{2}$$

33 kvadratini chiqarish.

$$x=\frac{-33±\sqrt{1089-4\left(-6\right)}}{2}$$

-4 ni -6 marotabaga ko'paytirish.

$$x=\frac{-33±\sqrt{1089+24}}{2}$$

1089 ni 24 ga qo'shish.

$$x=\frac{-33±\sqrt{1113}}{2}$$

x=\frac{-33±\sqrt{1113}}{2} tenglamasini yeching, bunda ± musbat. -33 ni \sqrt{1113} ga qo'shish.

$$x=\frac{\sqrt{1113}-33}{2}$$

x=\frac{-33±\sqrt{1113}}{2} tenglamasini yeching, bunda ± manfiy. -33 dan \sqrt{1113} ni ayirish.

$$x=\frac{-\sqrt{1113}-33}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{1113}-33}{2} x=\frac{-\sqrt{1113}-33}{2}$$

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.

$$x^{2}+33x=6$$

33 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{33}{2} olish uchun. Keyin, \frac{33}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+33x+\left(\frac{33}{2}\right)^{2}=6+\left(\frac{33}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{33}{2} kvadratini chiqarish.

$$x^{2}+33x+\frac{1089}{4}=6+\frac{1089}{4}$$

6 ni \frac{1089}{4} ga qo'shish.

$$x^{2}+33x+\frac{1089}{4}=\frac{1113}{4}$$

x^{2}+33x+\frac{1089}{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.

$$\left(x+\frac{33}{2}\right)^{2}=\frac{1113}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{33}{2}\right)^{2}}=\sqrt{\frac{1113}{4}}$$

Qisqartirish.

$$x+\frac{33}{2}=\frac{\sqrt{1113}}{2} x+\frac{33}{2}=-\frac{\sqrt{1113}}{2}$$

Tenglamaning ikkala tarafidan \frac{33}{2} ni ayirish.

$$x=\frac{\sqrt{1113}-33}{2} x=\frac{-\sqrt{1113}-33}{2}$$