2x^2+6x-1=0 eching

$$2 x ^ { 2 } + 6 x - 1 = 0$$

x uchun yechish

$$x=\frac{\sqrt{11}-3}{2}\approx 0,158312395$$

$$x=\frac{-\sqrt{11}-3}{2}\approx -3,158312395$$

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.

$$2x^{2}+6x-1=0$$

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

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-1\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+8}}{2\times 2}$$

36 ni 8 ga qo'shish.

$$x=\frac{-6±\sqrt{44}}{2\times 2}$$

44 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{11}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{11}}{4}$$

x=\frac{-6±2\sqrt{11}}{4} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{11} ga qo'shish.

$$x=\frac{2\sqrt{11}-6}{4}$$

-6+2\sqrt{11} ni 4 ga bo'lish.

$$x=\frac{\sqrt{11}-3}{2}$$

x=\frac{-6±2\sqrt{11}}{4} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{11} ni ayirish.

$$x=\frac{-2\sqrt{11}-6}{4}$$

-6-2\sqrt{11} ni 4 ga bo'lish.

$$x=\frac{-\sqrt{11}-3}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{11}-3}{2} x=\frac{-\sqrt{11}-3}{2}$$

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.

$$2x^{2}+6x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}+6x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$2x^{2}+6x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$2x^{2}+6x=1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+6x}{2}=\frac{1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{2}x=\frac{1}{2}$$

6 ni 2 ga bo'lish.

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

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

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+3x+\frac{9}{4}=\frac{11}{4}$$

x^{2}+3x+\frac{9}{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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{11}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{11}}{2} x+\frac{3}{2}=-\frac{\sqrt{11}}{2}$$

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

$$x=\frac{\sqrt{11}-3}{2} x=\frac{-\sqrt{11}-3}{2}$$