5x^2+25x+4=0 eching

$$5 x ^ { 2 } + 25 x + 4 = 0$$

x uchun yechish

$$x=\frac{\sqrt{545}}{10}-\frac{5}{2}\approx -0,165476494$$

$$x=-\frac{\sqrt{545}}{10}-\frac{5}{2}\approx -4,834523506$$

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.

$$5x^{2}+25x+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 25 ni b va 4 ni c bilan almashtiring.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5\times 4}}{2\times 5}$$

25 kvadratini chiqarish.

$$x=\frac{-25±\sqrt{625-4\times 5\times 4}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-20\times 4}}{2\times 5}$$

-20 ni 4 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-80}}{2\times 5}$$

625 ni -80 ga qo'shish.

$$x=\frac{-25±\sqrt{545}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{545}}{10}$$

x=\frac{-25±\sqrt{545}}{10} tenglamasini yeching, bunda ± musbat. -25 ni \sqrt{545} ga qo'shish.

$$x=\frac{\sqrt{545}-25}{10}$$

-25+\sqrt{545} ni 10 ga bo'lish.

$$x=\frac{\sqrt{545}}{10}-\frac{5}{2}$$

x=\frac{-25±\sqrt{545}}{10} tenglamasini yeching, bunda ± manfiy. -25 dan \sqrt{545} ni ayirish.

$$x=\frac{-\sqrt{545}-25}{10}$$

-25-\sqrt{545} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{545}}{10}-\frac{5}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{545}}{10}-\frac{5}{2} x=-\frac{\sqrt{545}}{10}-\frac{5}{2}$$

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.

$$5x^{2}+25x+4=0$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$5x^{2}+25x+4-4=-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$5x^{2}+25x=-4$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+25x}{5}=-\frac{4}{5}$$

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

$$x^{2}+\frac{25}{5}x=-\frac{4}{5}$$

25 ni 5 ga bo'lish.

$$x^{2}+5x=-\frac{4}{5}$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{4}{5}+\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$x^{2}+5x+\frac{25}{4}=-\frac{4}{5}+\frac{25}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{4}{5} ni \frac{25}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+5x+\frac{25}{4}=\frac{109}{20}$$

x^{2}+5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{109}{20}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{109}{20}}$$

Qisqartirish.

$$x+\frac{5}{2}=\frac{\sqrt{545}}{10} x+\frac{5}{2}=-\frac{\sqrt{545}}{10}$$

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

$$x=\frac{\sqrt{545}}{10}-\frac{5}{2} x=-\frac{\sqrt{545}}{10}-\frac{5}{2}$$