5x^2+6x-9=0 eching

$$5 x ^ { 2 } + 6 x - 9 = 0$$

x uchun yechish

$$x=\frac{3\sqrt{6}-3}{5}\approx 0,869693846$$

$$x=\frac{-3\sqrt{6}-3}{5}\approx -2,069693846$$

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.

$$5x^{2}+6x-9=0$$

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 -9 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-9\right)}}{2\times 5}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-9\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-20\left(-9\right)}}{2\times 5}$$

-20 ni -9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+180}}{2\times 5}$$

36 ni 180 ga qo'shish.

$$x=\frac{-6±\sqrt{216}}{2\times 5}$$

216 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6\sqrt{6}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±6\sqrt{6}}{10}$$

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

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

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

$$x=\frac{3\sqrt{6}-3}{5}$$

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

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

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

$$x=\frac{-3\sqrt{6}-3}{5}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{6}-3}{5} x=\frac{-3\sqrt{6}-3}{5}$$

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.

$$5x^{2}+6x-9=0$$

9 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+6x-9-\left(-9\right)=-\left(-9\right)$$

O‘zidan -9 ayirilsa 0 qoladi.

$$5x^{2}+6x=-\left(-9\right)$$

0 dan -9 ni ayirish.

$$5x^{2}+6x=9$$

Ikki tarafini 5 ga bo‘ling.

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

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

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

\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.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{9}{5}+\left(\frac{3}{5}\right)^{2}$$

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

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

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

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{54}{25}$$

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.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{54}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{54}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{3\sqrt{6}}{5} x+\frac{3}{5}=-\frac{3\sqrt{6}}{5}$$

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

$$x=\frac{3\sqrt{6}-3}{5} x=\frac{-3\sqrt{6}-3}{5}$$