4x(2x+1)=0 eching

$$4 x ( 2 x + 1 ) = 0$$

x uchun yechish

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

$$x=0$$

Faktorlash qadamlari

4x ga 2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$8x^{2}+4x=0$$

x omili.

$$x\left(8x+4\right)=0$$

Tenglamani yechish uchun x=0 va 8x+4=0 ni yeching.

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

Kvadrat tenglamalaridan foydalanish qoidalari

4x ga 2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$8x^{2}+4x=0$$

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

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 8}$$

4^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{-4±4}{16}$$

x=\frac{-4±4}{16} tenglamasini yeching, bunda ± musbat. -4 ni 4 ga qo'shish.

$$x=\frac{0}{16}$$

0 ni 16 ga bo'lish.

$$x=0$$

x=\frac{-4±4}{16} tenglamasini yeching, bunda ± manfiy. -4 dan 4 ni ayirish.

$$x=-\frac{8}{16}$$

\frac{-8}{16} ulushini 8 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

4x ga 2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$8x^{2}+4x=0$$

Ikki tarafini 8 ga bo‘ling.

$$\frac{8x^{2}+4x}{8}=\frac{0}{8}$$

8 ga bo'lish 8 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{8}x=\frac{0}{8}$$

\frac{4}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{1}{2}x=\frac{0}{8}$$

0 ni 8 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.

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