7x^2+5x=0 eching

$$7 x ^ { 2 } + 5 x = 0$$

x uchun yechish

$$x=-\frac{5}{7}\approx -0,714285714$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(7x+5\right)=0$$

Tenglamani yechish uchun x=0 va 7x+5=0 ni yeching.

$$x=0 x=-\frac{5}{7}$$

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.

$$7x^{2}+5x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 7 ni a, 5 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 7}$$

5^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-5±5}{2\times 7}$$

2 ni 7 marotabaga ko'paytirish.

$$x=\frac{-5±5}{14}$$

x=\frac{-5±5}{14} tenglamasini yeching, bunda ± musbat. -5 ni 5 ga qo'shish.

$$x=\frac{0}{14}$$

0 ni 14 ga bo'lish.

$$x=0$$

x=\frac{-5±5}{14} tenglamasini yeching, bunda ± manfiy. -5 dan 5 ni ayirish.

$$x=-\frac{10}{14}$$

\frac{-10}{14} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{5}{7}$$

Tenglama yechildi.

$$x=0 x=-\frac{5}{7}$$

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.

$$7x^{2}+5x=0$$

Ikki tarafini 7 ga bo‘ling.

$$\frac{7x^{2}+5x}{7}=\frac{0}{7}$$

7 ga bo'lish 7 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{5}{7}x=\frac{0}{7}$$

0 ni 7 ga bo'lish.

$$x^{2}+\frac{5}{7}x=0$$

\frac{5}{7} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{14} olish uchun. Keyin, \frac{5}{14} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{7}x+\left(\frac{5}{14}\right)^{2}=\left(\frac{5}{14}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{14} kvadratini chiqarish.

$$x^{2}+\frac{5}{7}x+\frac{25}{196}=\frac{25}{196}$$

x^{2}+\frac{5}{7}x+\frac{25}{196} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{14}\right)^{2}=\frac{25}{196}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{14}\right)^{2}}=\sqrt{\frac{25}{196}}$$

Qisqartirish.

$$x+\frac{5}{14}=\frac{5}{14} x+\frac{5}{14}=-\frac{5}{14}$$

Tenglamaning ikkala tarafidan \frac{5}{14} ni ayirish.

$$x=0 x=-\frac{5}{7}$$