5x^2+x-7=0 eching

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

x uchun yechish

$$x = \frac{\sqrt{141} - 1}{10} \approx 1,087434209$$

$$x=\frac{-\sqrt{141}-1}{10}\approx -1,287434209$$

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 1 ni b va -7 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 5\left(-7\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-20\left(-7\right)}}{2\times 5}$$

-20 ni -7 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+140}}{2\times 5}$$

1 ni 140 ga qo'shish.

$$x=\frac{-1±\sqrt{141}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{141}}{10}$$

x=\frac{-1±\sqrt{141}}{10} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{141} ga qo'shish.

$$x=\frac{\sqrt{141}-1}{10}$$

x=\frac{-1±\sqrt{141}}{10} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{141} ni ayirish.

$$x=\frac{-\sqrt{141}-1}{10}$$

Tenglama yechildi.

$$x=\frac{\sqrt{141}-1}{10} x=\frac{-\sqrt{141}-1}{10}$$

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.

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

7 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -7 ayirilsa 0 qoladi.

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

0 dan -7 ni ayirish.

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

Ikki tarafini 5 ga bo‘ling.

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

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

$$x^{2}+\frac{1}{5}x=\frac{7}{5}$$

\frac{1}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{10} olish uchun. Keyin, \frac{1}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{7}{5}+\left(\frac{1}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{10} kvadratini chiqarish.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{7}{5}+\frac{1}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{5} ni \frac{1}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{141}{100}$$

x^{2}+\frac{1}{5}x+\frac{1}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{141}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{141}{100}}$$

Qisqartirish.

$$x+\frac{1}{10}=\frac{\sqrt{141}}{10} x+\frac{1}{10}=-\frac{\sqrt{141}}{10}$$

Tenglamaning ikkala tarafidan \frac{1}{10} ni ayirish.

$$x=\frac{\sqrt{141}-1}{10} x=\frac{-\sqrt{141}-1}{10}$$