5x^2+12x-7=0 eching

$$5 { x }^{ 2 } +12x-7=0$$

x uchun yechish

$$x=\frac{\sqrt{71}-6}{5}\approx 0,485229955$$

$$x=\frac{-\sqrt{71}-6}{5}\approx -2,885229955$$

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.

$$5x^{2}+12x-7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 12 ni b va -7 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-7\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-20\left(-7\right)}}{2\times 5}$$

-20 ni -7 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+140}}{2\times 5}$$

144 ni 140 ga qo'shish.

$$x=\frac{-12±\sqrt{284}}{2\times 5}$$

284 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±2\sqrt{71}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-12±2\sqrt{71}}{10}$$

x=\frac{-12±2\sqrt{71}}{10} tenglamasini yeching, bunda ± musbat. -12 ni 2\sqrt{71} ga qo'shish.

$$x=\frac{2\sqrt{71}-12}{10}$$

-12+2\sqrt{71} ni 10 ga bo'lish.

$$x=\frac{\sqrt{71}-6}{5}$$

x=\frac{-12±2\sqrt{71}}{10} tenglamasini yeching, bunda ± manfiy. -12 dan 2\sqrt{71} ni ayirish.

$$x=\frac{-2\sqrt{71}-12}{10}$$

-12-2\sqrt{71} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{71}-6}{5}$$

Tenglama yechildi.

$$x=\frac{\sqrt{71}-6}{5} x=\frac{-\sqrt{71}-6}{5}$$

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.

$$5x^{2}+12x-7=0$$

7 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+12x-7-\left(-7\right)=-\left(-7\right)$$

O‘zidan -7 ayirilsa 0 qoladi.

$$5x^{2}+12x=-\left(-7\right)$$

0 dan -7 ni ayirish.

$$5x^{2}+12x=7$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+12x}{5}=\frac{7}{5}$$

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

$$x^{2}+\frac{12}{5}x=\frac{7}{5}$$

\frac{12}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{6}{5} olish uchun. Keyin, \frac{6}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=\frac{7}{5}+\left(\frac{6}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{6}{5} kvadratini chiqarish.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{7}{5}+\frac{36}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{5} ni \frac{36}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{71}{25}$$

x^{2}+\frac{12}{5}x+\frac{36}{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.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{71}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{71}{25}}$$

Qisqartirish.

$$x+\frac{6}{5}=\frac{\sqrt{71}}{5} x+\frac{6}{5}=-\frac{\sqrt{71}}{5}$$

Tenglamaning ikkala tarafidan \frac{6}{5} ni ayirish.

$$x=\frac{\sqrt{71}-6}{5} x=\frac{-\sqrt{71}-6}{5}$$