10x^2-5x=0 eching

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

x uchun yechish

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

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(10x-5\right)=0$$

Tenglamani yechish uchun x=0 va 10x-5=0 ni yeching.

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

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.

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

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

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

\left(-5\right)^{2} ning kvadrat ildizini chiqarish.

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

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±5}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{5±5}{20}$$

x=\frac{5±5}{20} tenglamasini yeching, bunda ± musbat. 5 ni 5 ga qo'shish.

$$x=\frac{10}{20}$$

\frac{10}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{2}$$

x=\frac{5±5}{20} tenglamasini yeching, bunda ± manfiy. 5 dan 5 ni ayirish.

$$x=\frac{0}{20}$$

0 ni 20 ga bo'lish.

$$x=0$$

Tenglama yechildi.

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

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.

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

Ikki tarafini 10 ga bo‘ling.

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

10 ga bo'lish 10 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{5}{10}\right)x=\frac{0}{10}$$

\frac{-5}{10} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{2}x=\frac{0}{10}$$

0 ni 10 ga bo'lish.

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

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

x^{2}-\frac{1}{2}x+\frac{1}{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.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

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