3x^2+4x-1=0 eching

$$3 x ^ { 2 } + 4 x - 1 = 0$$

x uchun yechish

$$x=\frac{\sqrt{7}-2}{3}\approx 0,215250437$$

$$x=\frac{-\sqrt{7}-2}{3}\approx -1,54858377$$

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.

$$3x^{2}+4x-1=0$$

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

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-1\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-12\left(-1\right)}}{2\times 3}$$

-12 ni -1 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+12}}{2\times 3}$$

16 ni 12 ga qo'shish.

$$x=\frac{-4±\sqrt{28}}{2\times 3}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±2\sqrt{7}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-4±2\sqrt{7}}{6}$$

x=\frac{-4±2\sqrt{7}}{6} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}-4}{6}$$

-4+2\sqrt{7} ni 6 ga bo'lish.

$$x=\frac{\sqrt{7}-2}{3}$$

x=\frac{-4±2\sqrt{7}}{6} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{7} ni ayirish.

$$x=\frac{-2\sqrt{7}-4}{6}$$

-4-2\sqrt{7} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{7}-2}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$

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.

$$3x^{2}+4x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$3x^{2}+4x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$3x^{2}+4x=-\left(-1\right)$$

0 dan -1 ni ayirish.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{3} olish uchun. Keyin, \frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{1}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{3}+\frac{4}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{3} ni \frac{4}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{9}$$

x^{2}+\frac{4}{3}x+\frac{4}{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.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{7}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Qisqartirish.

$$x+\frac{2}{3}=\frac{\sqrt{7}}{3} x+\frac{2}{3}=-\frac{\sqrt{7}}{3}$$

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

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$