5x^2+15x=12x-13 eching

$$5 { x }^{ 2 } +15x=12x-13$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{251}i}{10}\approx -0,3+1,584297952i$$

$$x=\frac{-\sqrt{251}i-3}{10}\approx -0,3-1,584297952i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 12x ni ayirish.

$$5x^{2}+15x-12x=-13$$

3x ni olish uchun 15x va -12x ni birlashtirish.

$$5x^{2}+3x=-13$$

13 ni ikki tarafga qo’shing.

$$5x^{2}+3x+13=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 3 ni b va 13 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\times 13}}{2\times 5}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times 5\times 13}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-20\times 13}}{2\times 5}$$

-20 ni 13 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-260}}{2\times 5}$$

9 ni -260 ga qo'shish.

$$x=\frac{-3±\sqrt{-251}}{2\times 5}$$

-251 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±\sqrt{251}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{251}i}{10}$$

x=\frac{-3±\sqrt{251}i}{10} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{251} ga qo'shish.

$$x=\frac{-3+\sqrt{251}i}{10}$$

x=\frac{-3±\sqrt{251}i}{10} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{251} ni ayirish.

$$x=\frac{-\sqrt{251}i-3}{10}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{251}i}{10} x=\frac{-\sqrt{251}i-3}{10}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 12x ni ayirish.

$$5x^{2}+15x-12x=-13$$

3x ni olish uchun 15x va -12x ni birlashtirish.

$$5x^{2}+3x=-13$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+3x}{5}=-\frac{13}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{3}{5}x=-\frac{13}{5}$$

\frac{3}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{10} olish uchun. Keyin, \frac{3}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=-\frac{13}{5}+\left(\frac{3}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{10} kvadratini chiqarish.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{13}{5}+\frac{9}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{13}{5} ni \frac{9}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{251}{100}$$

x^{2}+\frac{3}{5}x+\frac{9}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{10}\right)^{2}=-\frac{251}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{-\frac{251}{100}}$$

Qisqartirish.

$$x+\frac{3}{10}=\frac{\sqrt{251}i}{10} x+\frac{3}{10}=-\frac{\sqrt{251}i}{10}$$

Tenglamaning ikkala tarafidan \frac{3}{10} ni ayirish.

$$x=\frac{-3+\sqrt{251}i}{10} x=\frac{-\sqrt{251}i-3}{10}$$