3x^2+3x-2=0 eching

$$3 x ^ { 2 } + 3 x - 2 = 0$$

x uchun yechish

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}\approx 0,457427108$$

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}\approx -1,457427108$$

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.

$$3x^{2}+3x-2=0$$

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

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

3 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

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

-12 ni -2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+24}}{2\times 3}$$

9 ni 24 ga qo'shish.

$$x=\frac{-3±\sqrt{33}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{33}}{6}$$

x=\frac{-3±\sqrt{33}}{6} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{33} ga qo'shish.

$$x=\frac{\sqrt{33}-3}{6}$$

-3+\sqrt{33} ni 6 ga bo'lish.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}$$

x=\frac{-3±\sqrt{33}}{6} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{33} ni ayirish.

$$x=\frac{-\sqrt{33}-3}{6}$$

-3-\sqrt{33} ni 6 ga bo'lish.

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2} x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

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.

$$3x^{2}+3x-2=0$$

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

0 dan -2 ni ayirish.

$$3x^{2}+3x=2$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+3x}{3}=\frac{2}{3}$$

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

$$x^{2}+\frac{3}{3}x=\frac{2}{3}$$

3 ni 3 ga bo'lish.

$$x^{2}+x=\frac{2}{3}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{2}{3}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{2}{3}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2}{3} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+x+\frac{1}{4}=\frac{11}{12}$$

x^{2}+x+\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.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{11}{12}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{11}{12}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{33}}{6} x+\frac{1}{2}=-\frac{\sqrt{33}}{6}$$

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

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2} x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$