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

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

x uchun yechish

$$x=\frac{\sqrt{10}-1}{3}\approx 0,72075922$$

$$x=\frac{-\sqrt{10}-1}{3}\approx -1,387425887$$

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.

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

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

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

2 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

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

-12 ni -3 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+36}}{2\times 3}$$

4 ni 36 ga qo'shish.

$$x=\frac{-2±\sqrt{40}}{2\times 3}$$

40 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{10}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±2\sqrt{10}}{6}$$

x=\frac{-2±2\sqrt{10}}{6} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{10} ga qo'shish.

$$x=\frac{2\sqrt{10}-2}{6}$$

-2+2\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{\sqrt{10}-1}{3}$$

x=\frac{-2±2\sqrt{10}}{6} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{10} ni ayirish.

$$x=\frac{-2\sqrt{10}-2}{6}$$

-2-2\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{10}-1}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{10}-1}{3} x=\frac{-\sqrt{10}-1}{3}$$

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.

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

3 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -3 ayirilsa 0 qoladi.

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

0 dan -3 ni ayirish.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

3 ni 3 ga bo'lish.

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

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=1+\frac{1}{9}$$

1 ni \frac{1}{9} ga qo'shish.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{10}{9}$$

x^{2}+\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{10}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{\sqrt{10}}{3} x+\frac{1}{3}=-\frac{\sqrt{10}}{3}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

$$x=\frac{\sqrt{10}-1}{3} x=\frac{-\sqrt{10}-1}{3}$$