5x^2+8x=-2 eching

$$5 { x }^{ 2 } +8x=-2$$

x uchun yechish

$$x=\frac{\sqrt{6}-4}{5}\approx -0,310102051$$

$$x=\frac{-\sqrt{6}-4}{5}\approx -1,289897949$$

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.

$$5x^{2}+8x=-2$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+8x-\left(-2\right)=-2-\left(-2\right)$$

O‘zidan -2 ayirilsa 0 qoladi.

$$5x^{2}+8x-\left(-2\right)=0$$

0 dan -2 ni ayirish.

$$5x^{2}+8x+2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 8 ni b va 2 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 5\times 2}}{2\times 5}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 5\times 2}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-20\times 2}}{2\times 5}$$

-20 ni 2 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-40}}{2\times 5}$$

64 ni -40 ga qo'shish.

$$x=\frac{-8±\sqrt{24}}{2\times 5}$$

24 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{6}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-8±2\sqrt{6}}{10}$$

x=\frac{-8±2\sqrt{6}}{10} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{6} ga qo'shish.

$$x=\frac{2\sqrt{6}-8}{10}$$

-8+2\sqrt{6} ni 10 ga bo'lish.

$$x=\frac{\sqrt{6}-4}{5}$$

x=\frac{-8±2\sqrt{6}}{10} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{6} ni ayirish.

$$x=\frac{-2\sqrt{6}-8}{10}$$

-8-2\sqrt{6} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{6}-4}{5}$$

Tenglama yechildi.

$$x=\frac{\sqrt{6}-4}{5} x=\frac{-\sqrt{6}-4}{5}$$

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.

$$5x^{2}+8x=-2$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+8x}{5}=-\frac{2}{5}$$

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

$$x^{2}+\frac{8}{5}x=-\frac{2}{5}$$

\frac{8}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{4}{5} olish uchun. Keyin, \frac{4}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=-\frac{2}{5}+\left(\frac{4}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{4}{5} kvadratini chiqarish.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=-\frac{2}{5}+\frac{16}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{2}{5} ni \frac{16}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{6}{25}$$

x^{2}+\frac{8}{5}x+\frac{16}{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.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{6}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{6}{25}}$$

Qisqartirish.

$$x+\frac{4}{5}=\frac{\sqrt{6}}{5} x+\frac{4}{5}=-\frac{\sqrt{6}}{5}$$

Tenglamaning ikkala tarafidan \frac{4}{5} ni ayirish.

$$x=\frac{\sqrt{6}-4}{5} x=\frac{-\sqrt{6}-4}{5}$$