-3/3x^2+16/5x-1=0 eching

$$- \frac { 3 } { 3 } x ^ { 2 } + \frac { 16 } { 5 } x - 1 = 0$$

x uchun yechish

$$x = \frac{\sqrt{39} + 8}{5} \approx 2,8489996$$

$$x=\frac{8-\sqrt{39}}{5}\approx 0,3510004$$

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.

$$-x^{2}+\frac{16}{5}x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, \frac{16}{5} ni b va -1 ni c bilan almashtiring.

$$x=\frac{-\frac{16}{5}±\sqrt{\left(\frac{16}{5}\right)^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{16}{5} kvadratini chiqarish.

$$x=\frac{-\frac{16}{5}±\sqrt{\frac{256}{25}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\frac{16}{5}±\sqrt{\frac{256}{25}+4\left(-1\right)}}{2\left(-1\right)}$$

4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\frac{16}{5}±\sqrt{\frac{256}{25}-4}}{2\left(-1\right)}$$

\frac{256}{25} ni -4 ga qo'shish.

$$x=\frac{-\frac{16}{5}±\sqrt{\frac{156}{25}}}{2\left(-1\right)}$$

\frac{156}{25} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{16}{5}±\frac{2\sqrt{39}}{5}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\frac{16}{5}±\frac{2\sqrt{39}}{5}}{-2}$$

x=\frac{-\frac{16}{5}±\frac{2\sqrt{39}}{5}}{-2} tenglamasini yeching, bunda ± musbat. -\frac{16}{5} ni \frac{2\sqrt{39}}{5} ga qo'shish.

$$x=\frac{2\sqrt{39}-16}{-2\times 5}$$

\frac{-16+2\sqrt{39}}{5} ni -2 ga bo'lish.

$$x=\frac{8-\sqrt{39}}{5}$$

x=\frac{-\frac{16}{5}±\frac{2\sqrt{39}}{5}}{-2} tenglamasini yeching, bunda ± manfiy. -\frac{16}{5} dan \frac{2\sqrt{39}}{5} ni ayirish.

$$x=\frac{-2\sqrt{39}-16}{-2\times 5}$$

\frac{-16-2\sqrt{39}}{5} ni -2 ga bo'lish.

$$x=\frac{\sqrt{39}+8}{5}$$

Tenglama yechildi.

$$x=\frac{8-\sqrt{39}}{5} x=\frac{\sqrt{39}+8}{5}$$

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.

$$-x^{2}+\frac{16}{5}x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$-x^{2}+\frac{16}{5}x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$-x^{2}+\frac{16}{5}x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$-x^{2}+\frac{16}{5}x=1$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}+\frac{16}{5}x}{-1}=\frac{1}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{\frac{16}{5}}{-1}x=\frac{1}{-1}$$

\frac{16}{5} ni -1 ga bo'lish.

$$x^{2}-\frac{16}{5}x=\frac{1}{-1}$$

1 ni -1 ga bo'lish.

$$x^{2}-\frac{16}{5}x=-1$$

-\frac{16}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{8}{5} olish uchun. Keyin, -\frac{8}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{16}{5}x+\left(-\frac{8}{5}\right)^{2}=-1+\left(-\frac{8}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{8}{5} kvadratini chiqarish.

$$x^{2}-\frac{16}{5}x+\frac{64}{25}=-1+\frac{64}{25}$$

-1 ni \frac{64}{25} ga qo'shish.

$$x^{2}-\frac{16}{5}x+\frac{64}{25}=\frac{39}{25}$$

x^{2}-\frac{16}{5}x+\frac{64}{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.

$$\left(x-\frac{8}{5}\right)^{2}=\frac{39}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{8}{5}\right)^{2}}=\sqrt{\frac{39}{25}}$$

Qisqartirish.

$$x-\frac{8}{5}=\frac{\sqrt{39}}{5} x-\frac{8}{5}=-\frac{\sqrt{39}}{5}$$

\frac{8}{5} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{39}+8}{5} x=\frac{8-\sqrt{39}}{5}$$