5x^2-10x-2=0 eching

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

x uchun yechish

$$x=\frac{\sqrt{35}}{5}+1\approx 2,183215957$$

$$x=-\frac{\sqrt{35}}{5}+1\approx -0,183215957$$

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.

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

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 -2 ni c bilan almashtiring.

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

-10 kvadratini chiqarish.

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

-4 ni 5 marotabaga ko'paytirish.

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

-20 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+40}}{2\times 5}$$

100 ni 40 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{140}}{2\times 5}$$

140 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{35}}{2\times 5}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{35}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{10±2\sqrt{35}}{10}$$

x=\frac{10±2\sqrt{35}}{10} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{35} ga qo'shish.

$$x=\frac{2\sqrt{35}+10}{10}$$

10+2\sqrt{35} ni 10 ga bo'lish.

$$x=\frac{\sqrt{35}}{5}+1$$

x=\frac{10±2\sqrt{35}}{10} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{35} ni ayirish.

$$x=\frac{10-2\sqrt{35}}{10}$$

10-2\sqrt{35} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{35}}{5}+1$$

Tenglama yechildi.

$$x=\frac{\sqrt{35}}{5}+1 x=-\frac{\sqrt{35}}{5}+1$$

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.

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

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

0 dan -2 ni ayirish.

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

Ikki tarafini 5 ga bo‘ling.

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

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

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

-10 ni 5 ga bo'lish.

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

-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.

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

\frac{2}{5} ni 1 ga qo'shish.

$$x^{2}-2x+1=\frac{7}{5}$$

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.

$$\left(x-1\right)^{2}=\frac{7}{5}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{7}{5}}$$

Qisqartirish.

$$x-1=\frac{\sqrt{35}}{5} x-1=-\frac{\sqrt{35}}{5}$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{35}}{5}+1 x=-\frac{\sqrt{35}}{5}+1$$