5x^2+3x=17 eching

$$5 x ^ { 2 } + 3 x = 17$$

x uchun yechish

$$x = \frac{\sqrt{349} - 3}{10} \approx 1,568154169$$

$$x=\frac{-\sqrt{349}-3}{10}\approx -2,168154169$$

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.

$$5x^{2}+3x=17$$

Tenglamaning ikkala tarafidan 17 ni ayirish.

$$5x^{2}+3x-17=17-17$$

O‘zidan 17 ayirilsa 0 qoladi.

$$5x^{2}+3x-17=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 3 ni b va -17 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-17\right)}}{2\times 5}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times 5\left(-17\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-20\left(-17\right)}}{2\times 5}$$

-20 ni -17 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+340}}{2\times 5}$$

9 ni 340 ga qo'shish.

$$x=\frac{-3±\sqrt{349}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{349}}{10}$$

x=\frac{-3±\sqrt{349}}{10} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{349} ga qo'shish.

$$x=\frac{\sqrt{349}-3}{10}$$

x=\frac{-3±\sqrt{349}}{10} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{349} ni ayirish.

$$x=\frac{-\sqrt{349}-3}{10}$$

Tenglama yechildi.

$$x=\frac{\sqrt{349}-3}{10} x=\frac{-\sqrt{349}-3}{10}$$

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.

$$5x^{2}+3x=17$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+3x}{5}=\frac{17}{5}$$

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

$$x^{2}+\frac{3}{5}x=\frac{17}{5}$$

\frac{3}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{10} olish uchun. Keyin, \frac{3}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\frac{17}{5}+\left(\frac{3}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{10} kvadratini chiqarish.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{17}{5}+\frac{9}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{17}{5} ni \frac{9}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{349}{100}$$

x^{2}+\frac{3}{5}x+\frac{9}{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.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{349}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{349}{100}}$$

Qisqartirish.

$$x+\frac{3}{10}=\frac{\sqrt{349}}{10} x+\frac{3}{10}=-\frac{\sqrt{349}}{10}$$

Tenglamaning ikkala tarafidan \frac{3}{10} ni ayirish.

$$x=\frac{\sqrt{349}-3}{10} x=\frac{-\sqrt{349}-3}{10}$$