5x^2+6x-1=0 eching

$$5 x ^ { 2 } + 6 x - 1 = 0$$

x uchun yechish

$$x=\frac{\sqrt{14}-3}{5}\approx 0,148331477$$

$$x=\frac{-\sqrt{14}-3}{5}\approx -1,348331477$$

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.

$$5x^{2}+6x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 6 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-1\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-20\left(-1\right)}}{2\times 5}$$

-20 ni -1 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+20}}{2\times 5}$$

36 ni 20 ga qo'shish.

$$x=\frac{-6±\sqrt{56}}{2\times 5}$$

56 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{14}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{14}}{10}$$

x=\frac{-6±2\sqrt{14}}{10} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{14} ga qo'shish.

$$x=\frac{2\sqrt{14}-6}{10}$$

-6+2\sqrt{14} ni 10 ga bo'lish.

$$x=\frac{\sqrt{14}-3}{5}$$

x=\frac{-6±2\sqrt{14}}{10} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{14} ni ayirish.

$$x=\frac{-2\sqrt{14}-6}{10}$$

-6-2\sqrt{14} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{14}-3}{5}$$

Tenglama yechildi.

$$x=\frac{\sqrt{14}-3}{5} x=\frac{-\sqrt{14}-3}{5}$$

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.

$$5x^{2}+6x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+6x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$5x^{2}+6x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$5x^{2}+6x=1$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+6x}{5}=\frac{1}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{5}x=\frac{1}{5}$$

\frac{6}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{5} olish uchun. Keyin, \frac{3}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{1}{5}+\left(\frac{3}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{5} kvadratini chiqarish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{1}{5}+\frac{9}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{5} ni \frac{9}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{14}{25}$$

x^{2}+\frac{6}{5}x+\frac{9}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{14}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{14}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{\sqrt{14}}{5} x+\frac{3}{5}=-\frac{\sqrt{14}}{5}$$

Tenglamaning ikkala tarafidan \frac{3}{5} ni ayirish.

$$x=\frac{\sqrt{14}-3}{5} x=\frac{-\sqrt{14}-3}{5}$$