5x^2-40x+85=0 eching

$$5 x ^ { 2 } - 40 x + 85 = 0$$

x uchun yechish (complex solution)

$$x=4+i$$

$$x=4-i$$

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.

$$5x^{2}-40x+85=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -40 ni b va 85 ni c bilan almashtiring.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\times 85}}{2\times 5}$$

-40 kvadratini chiqarish.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\times 85}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\times 85}}{2\times 5}$$

-20 ni 85 marotabaga ko'paytirish.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1700}}{2\times 5}$$

1600 ni -1700 ga qo'shish.

$$x=\frac{-\left(-40\right)±\sqrt{-100}}{2\times 5}$$

-100 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-40\right)±10i}{2\times 5}$$

-40 ning teskarisi 40 ga teng.

$$x=\frac{40±10i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{40±10i}{10}$$

x=\frac{40±10i}{10} tenglamasini yeching, bunda ± musbat. 40 ni 10i ga qo'shish.

$$x=\frac{40+10i}{10}$$

40+10i ni 10 ga bo'lish.

$$x=4+i$$

x=\frac{40±10i}{10} tenglamasini yeching, bunda ± manfiy. 40 dan 10i ni ayirish.

$$x=\frac{40-10i}{10}$$

40-10i ni 10 ga bo'lish.

$$x=4-i$$

Tenglama yechildi.

$$x=4+i x=4-i$$

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.

$$5x^{2}-40x+85=0$$

Tenglamaning ikkala tarafidan 85 ni ayirish.

$$5x^{2}-40x+85-85=-85$$

O‘zidan 85 ayirilsa 0 qoladi.

$$5x^{2}-40x=-85$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-40x}{5}=-\frac{85}{5}$$

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

$$x^{2}+\left(-\frac{40}{5}\right)x=-\frac{85}{5}$$

-40 ni 5 ga bo'lish.

$$x^{2}-8x=-\frac{85}{5}$$

-85 ni 5 ga bo'lish.

$$x^{2}-8x=-17$$

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

$$x^{2}-8x+\left(-4\right)^{2}=-17+\left(-4\right)^{2}$$

-4 kvadratini chiqarish.

$$x^{2}-8x+16=-17+16$$

-17 ni 16 ga qo'shish.

$$x^{2}-8x+16=-1$$

x^{2}-8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-4\right)^{2}=-1$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-4=i x-4=-i$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$x=4+i x=4-i$$