5x^2-25x-12=0 eching

$$5 { x }^{ 2 } -25x-12=0$$

x uchun yechish

$$x=\frac{\sqrt{865}}{10}+\frac{5}{2}\approx 5,441088234$$

$$x=-\frac{\sqrt{865}}{10}+\frac{5}{2}\approx -0,441088234$$

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.

$$5x^{2}-25x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -25 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

-25 kvadratini chiqarish.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 5\left(-12\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-25\right)±\sqrt{625-20\left(-12\right)}}{2\times 5}$$

-20 ni -12 marotabaga ko'paytirish.

$$x=\frac{-\left(-25\right)±\sqrt{625+240}}{2\times 5}$$

625 ni 240 ga qo'shish.

$$x=\frac{-\left(-25\right)±\sqrt{865}}{2\times 5}$$

-25 ning teskarisi 25 ga teng.

$$x=\frac{25±\sqrt{865}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{25±\sqrt{865}}{10}$$

x=\frac{25±\sqrt{865}}{10} tenglamasini yeching, bunda ± musbat. 25 ni \sqrt{865} ga qo'shish.

$$x=\frac{\sqrt{865}+25}{10}$$

25+\sqrt{865} ni 10 ga bo'lish.

$$x=\frac{\sqrt{865}}{10}+\frac{5}{2}$$

x=\frac{25±\sqrt{865}}{10} tenglamasini yeching, bunda ± manfiy. 25 dan \sqrt{865} ni ayirish.

$$x=\frac{25-\sqrt{865}}{10}$$

25-\sqrt{865} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{865}}{10}+\frac{5}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{865}}{10}+\frac{5}{2} x=-\frac{\sqrt{865}}{10}+\frac{5}{2}$$

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.

$$5x^{2}-25x-12=0$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}-25x-12-\left(-12\right)=-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$5x^{2}-25x=-\left(-12\right)$$

0 dan -12 ni ayirish.

$$5x^{2}-25x=12$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-25x}{5}=\frac{12}{5}$$

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

$$x^{2}+\left(-\frac{25}{5}\right)x=\frac{12}{5}$$

-25 ni 5 ga bo'lish.

$$x^{2}-5x=\frac{12}{5}$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\frac{12}{5}+\left(-\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{2} kvadratini chiqarish.

$$x^{2}-5x+\frac{25}{4}=\frac{12}{5}+\frac{25}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{12}{5} ni \frac{25}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-5x+\frac{25}{4}=\frac{173}{20}$$

x^{2}-5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{173}{20}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{173}{20}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{865}}{10} x-\frac{5}{2}=-\frac{\sqrt{865}}{10}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{865}}{10}+\frac{5}{2} x=-\frac{\sqrt{865}}{10}+\frac{5}{2}$$