5x^2-32x=48 eching

$$5 { x }^{ 2 } -32x=48$$

x uchun yechish

$$x = \frac{4 \sqrt{31} + 16}{5} \approx 7,65421149$$

$$x=\frac{16-4\sqrt{31}}{5}\approx -1,25421149$$

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.

$$5x^{2}-32x=48$$

Tenglamaning ikkala tarafidan 48 ni ayirish.

$$5x^{2}-32x-48=48-48$$

O‘zidan 48 ayirilsa 0 qoladi.

$$5x^{2}-32x-48=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -32 ni b va -48 ni c bilan almashtiring.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 5\left(-48\right)}}{2\times 5}$$

-32 kvadratini chiqarish.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 5\left(-48\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-32\right)±\sqrt{1024-20\left(-48\right)}}{2\times 5}$$

-20 ni -48 marotabaga ko'paytirish.

$$x=\frac{-\left(-32\right)±\sqrt{1024+960}}{2\times 5}$$

1024 ni 960 ga qo'shish.

$$x=\frac{-\left(-32\right)±\sqrt{1984}}{2\times 5}$$

1984 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-32\right)±8\sqrt{31}}{2\times 5}$$

-32 ning teskarisi 32 ga teng.

$$x=\frac{32±8\sqrt{31}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{32±8\sqrt{31}}{10}$$

x=\frac{32±8\sqrt{31}}{10} tenglamasini yeching, bunda ± musbat. 32 ni 8\sqrt{31} ga qo'shish.

$$x=\frac{8\sqrt{31}+32}{10}$$

32+8\sqrt{31} ni 10 ga bo'lish.

$$x=\frac{4\sqrt{31}+16}{5}$$

x=\frac{32±8\sqrt{31}}{10} tenglamasini yeching, bunda ± manfiy. 32 dan 8\sqrt{31} ni ayirish.

$$x=\frac{32-8\sqrt{31}}{10}$$

32-8\sqrt{31} ni 10 ga bo'lish.

$$x=\frac{16-4\sqrt{31}}{5}$$

Tenglama yechildi.

$$x=\frac{4\sqrt{31}+16}{5} x=\frac{16-4\sqrt{31}}{5}$$

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.

$$5x^{2}-32x=48$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-32x}{5}=\frac{48}{5}$$

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

$$x^{2}-\frac{32}{5}x=\frac{48}{5}$$

-\frac{32}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{16}{5} olish uchun. Keyin, -\frac{16}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{32}{5}x+\left(-\frac{16}{5}\right)^{2}=\frac{48}{5}+\left(-\frac{16}{5}\right)^{2}$$

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

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=\frac{48}{5}+\frac{256}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{48}{5} ni \frac{256}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=\frac{496}{25}$$

x^{2}-\frac{32}{5}x+\frac{256}{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.

$$\left(x-\frac{16}{5}\right)^{2}=\frac{496}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{16}{5}\right)^{2}}=\sqrt{\frac{496}{25}}$$

Qisqartirish.

$$x-\frac{16}{5}=\frac{4\sqrt{31}}{5} x-\frac{16}{5}=-\frac{4\sqrt{31}}{5}$$

\frac{16}{5} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{4\sqrt{31}+16}{5} x=\frac{16-4\sqrt{31}}{5}$$