2x^2+x=0 eching

$$2 x ^ { 2 } + x = 0$$

x uchun yechish

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

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(2x+1\right)=0$$

Tenglamani yechish uchun x=0 va 2x+1=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.

$$2x^{2}+x=0$$

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

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 2}$$

1^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-1±1}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±1}{4}$$

x=\frac{-1±1}{4} tenglamasini yeching, bunda ± musbat. -1 ni 1 ga qo'shish.

$$x=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$x=0$$

x=\frac{-1±1}{4} tenglamasini yeching, bunda ± manfiy. -1 dan 1 ni ayirish.

$$x=-\frac{2}{4}$$

\frac{-2}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

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

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.

$$2x^{2}+x=0$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+x}{2}=\frac{0}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

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

0 ni 2 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.

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

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

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