7=15x^2-5x eching

$$7 = 15 x ^ { 2 } - 5 x$$

x uchun yechish

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}\approx 0,869834104$$

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}\approx -0,53650077$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$15x^{2}-5x=7$$

Ikkala tarafdan 7 ni ayirish.

$$15x^{2}-5x-7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 15 ni a, -5 ni b va -7 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 15\left(-7\right)}}{2\times 15}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 15\left(-7\right)}}{2\times 15}$$

-4 ni 15 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-60\left(-7\right)}}{2\times 15}$$

-60 ni -7 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+420}}{2\times 15}$$

25 ni 420 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{445}}{2\times 15}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{445}}{2\times 15}$$

2 ni 15 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{445}}{30}$$

x=\frac{5±\sqrt{445}}{30} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{445} ga qo'shish.

$$x=\frac{\sqrt{445}+5}{30}$$

5+\sqrt{445} ni 30 ga bo'lish.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}$$

x=\frac{5±\sqrt{445}}{30} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{445} ni ayirish.

$$x=\frac{5-\sqrt{445}}{30}$$

5-\sqrt{445} ni 30 ga bo'lish.

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Tenglama yechildi.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$15x^{2}-5x=7$$

Ikki tarafini 15 ga bo‘ling.

$$\frac{15x^{2}-5x}{15}=\frac{7}{15}$$

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

$$x^{2}+\left(-\frac{5}{15}\right)x=\frac{7}{15}$$

\frac{-5}{15} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{3}x=\frac{7}{15}$$

-\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{6} olish uchun. Keyin, -\frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{7}{15}+\left(-\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{6} kvadratini chiqarish.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{7}{15}+\frac{1}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{15} ni \frac{1}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{89}{180}$$

x^{2}-\frac{1}{3}x+\frac{1}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{89}{180}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{89}{180}}$$

Qisqartirish.

$$x-\frac{1}{6}=\frac{\sqrt{445}}{30} x-\frac{1}{6}=-\frac{\sqrt{445}}{30}$$

\frac{1}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$