6x-15x^2=0 eching

$$6 x - 15 x ^ { 2 } = 0$$

x uchun yechish

$$x=\frac{2}{5}=0,4$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(6-15x\right)=0$$

Tenglamani yechish uchun x=0 va 6-15x=0 ni yeching.

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

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.

$$-15x^{2}+6x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -15 ni a, 6 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}}}{2\left(-15\right)}$$

6^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6}{2\left(-15\right)}$$

2 ni -15 marotabaga ko'paytirish.

$$x=\frac{-6±6}{-30}$$

x=\frac{-6±6}{-30} tenglamasini yeching, bunda ± musbat. -6 ni 6 ga qo'shish.

$$x=\frac{0}{-30}$$

0 ni -30 ga bo'lish.

$$x=0$$

x=\frac{-6±6}{-30} tenglamasini yeching, bunda ± manfiy. -6 dan 6 ni ayirish.

$$x=-\frac{12}{-30}$$

\frac{-12}{-30} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{2}{5}$$

Tenglama yechildi.

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

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.

$$-15x^{2}+6x=0$$

Ikki tarafini -15 ga bo‘ling.

$$\frac{-15x^{2}+6x}{-15}=\frac{0}{-15}$$

-15 ga bo'lish -15 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{-15}x=\frac{0}{-15}$$

\frac{6}{-15} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{2}{5}x=\frac{0}{-15}$$

0 ni -15 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{5} kvadratini chiqarish.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{1}{25}$$

x^{2}-\frac{2}{5}x+\frac{1}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{1}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Qisqartirish.

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

\frac{1}{5} ni tenglamaning ikkala tarafiga qo'shish.

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