5x^2+1=10x eching

$$5 { x }^{ 2 } +1=10x$$

x uchun yechish

$$x=\frac{2\sqrt{5}}{5}+1\approx 1,894427191$$

$$x=-\frac{2\sqrt{5}}{5}+1\approx 0,105572809$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 10x ni ayirish.

$$5x^{2}+1-10x=0$$

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.

$$5x^{2}-10x+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -10 ni b va 1 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5}}{2\times 5}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 5}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-20}}{2\times 5}$$

100 ni -20 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{80}}{2\times 5}$$

80 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±4\sqrt{5}}{2\times 5}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±4\sqrt{5}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{10±4\sqrt{5}}{10}$$

x=\frac{10±4\sqrt{5}}{10} tenglamasini yeching, bunda ± musbat. 10 ni 4\sqrt{5} ga qo'shish.

$$x=\frac{4\sqrt{5}+10}{10}$$

10+4\sqrt{5} ni 10 ga bo'lish.

$$x=\frac{2\sqrt{5}}{5}+1$$

x=\frac{10±4\sqrt{5}}{10} tenglamasini yeching, bunda ± manfiy. 10 dan 4\sqrt{5} ni ayirish.

$$x=\frac{10-4\sqrt{5}}{10}$$

10-4\sqrt{5} ni 10 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 10x ni ayirish.

$$5x^{2}+1-10x=0$$

Ikkala tarafdan 1 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$5x^{2}-10x=-1$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-10x}{5}=-\frac{1}{5}$$

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

$$x^{2}+\left(-\frac{10}{5}\right)x=-\frac{1}{5}$$

-10 ni 5 ga bo'lish.

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

-2 ni bo‘lish, x shartining koeffitsienti, 2 ga -1 olish uchun. Keyin, -1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

-\frac{1}{5} ni 1 ga qo'shish.

$$x^{2}-2x+1=\frac{4}{5}$$

x^{2}-2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-1=\frac{2\sqrt{5}}{5} x-1=-\frac{2\sqrt{5}}{5}$$

1 ni tenglamaning ikkala tarafiga qo'shish.

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