40x^2-8x=0 eching

$$40 x ^ { 2 } - 8 x = 0$$

x uchun yechish

$$x=\frac{1}{5}=0,2$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(40x-8\right)=0$$

Tenglamani yechish uchun x=0 va 40x-8=0 ni yeching.

$$x=0 x=\frac{1}{5}$$

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.

$$40x^{2}-8x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 40 ni a, -8 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 40}$$

\left(-8\right)^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±8}{2\times 40}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±8}{2\times 40}$$

2 ni 40 marotabaga ko'paytirish.

$$x=\frac{8±8}{80}$$

x=\frac{8±8}{80} tenglamasini yeching, bunda ± musbat. 8 ni 8 ga qo'shish.

$$x=\frac{16}{80}$$

\frac{16}{80} ulushini 16 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{5}$$

x=\frac{8±8}{80} tenglamasini yeching, bunda ± manfiy. 8 dan 8 ni ayirish.

$$x=\frac{0}{80}$$

0 ni 80 ga bo'lish.

$$x=0$$

Tenglama yechildi.

$$x=\frac{1}{5} x=0$$

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.

$$40x^{2}-8x=0$$

Ikki tarafini 40 ga bo‘ling.

$$\frac{40x^{2}-8x}{40}=\frac{0}{40}$$

40 ga bo'lish 40 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{8}{40}\right)x=\frac{0}{40}$$

\frac{-8}{40} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{5}x=\frac{0}{40}$$

0 ni 40 ga bo'lish.

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

-\frac{1}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{10} olish uchun. Keyin, -\frac{1}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{10} kvadratini chiqarish.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

x^{2}-\frac{1}{5}x+\frac{1}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Qisqartirish.

$$x-\frac{1}{10}=\frac{1}{10} x-\frac{1}{10}=-\frac{1}{10}$$

\frac{1}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{5} x=0$$