12x^2+25x-45=0 eching

$$12 { x }^{ 2 } +25x-45=0$$

x uchun yechish

$$x = \frac{\sqrt{2785} - 25}{24} \approx 1,157212467$$

$$x=\frac{-\sqrt{2785}-25}{24}\approx -3,2405458$$

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.

$$12x^{2}+25x-45=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 12 ni a, 25 ni b va -45 ni c bilan almashtiring.

$$x=\frac{-25±\sqrt{25^{2}-4\times 12\left(-45\right)}}{2\times 12}$$

25 kvadratini chiqarish.

$$x=\frac{-25±\sqrt{625-4\times 12\left(-45\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-48\left(-45\right)}}{2\times 12}$$

-48 ni -45 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625+2160}}{2\times 12}$$

625 ni 2160 ga qo'shish.

$$x=\frac{-25±\sqrt{2785}}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{2785}}{24}$$

x=\frac{-25±\sqrt{2785}}{24} tenglamasini yeching, bunda ± musbat. -25 ni \sqrt{2785} ga qo'shish.

$$x=\frac{\sqrt{2785}-25}{24}$$

x=\frac{-25±\sqrt{2785}}{24} tenglamasini yeching, bunda ± manfiy. -25 dan \sqrt{2785} ni ayirish.

$$x=\frac{-\sqrt{2785}-25}{24}$$

Tenglama yechildi.

$$x=\frac{\sqrt{2785}-25}{24} x=\frac{-\sqrt{2785}-25}{24}$$

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.

$$12x^{2}+25x-45=0$$

45 ni tenglamaning ikkala tarafiga qo'shish.

$$12x^{2}+25x-45-\left(-45\right)=-\left(-45\right)$$

O‘zidan -45 ayirilsa 0 qoladi.

$$12x^{2}+25x=-\left(-45\right)$$

0 dan -45 ni ayirish.

$$12x^{2}+25x=45$$

Ikki tarafini 12 ga bo‘ling.

$$\frac{12x^{2}+25x}{12}=\frac{45}{12}$$

12 ga bo'lish 12 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{25}{12}x=\frac{45}{12}$$

\frac{45}{12} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{25}{12}x=\frac{15}{4}$$

\frac{25}{12} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{25}{24} olish uchun. Keyin, \frac{25}{24} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{25}{12}x+\left(\frac{25}{24}\right)^{2}=\frac{15}{4}+\left(\frac{25}{24}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{25}{24} kvadratini chiqarish.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{15}{4}+\frac{625}{576}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{15}{4} ni \frac{625}{576} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{2785}{576}$$

x^{2}+\frac{25}{12}x+\frac{625}{576} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{25}{24}\right)^{2}=\frac{2785}{576}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{25}{24}\right)^{2}}=\sqrt{\frac{2785}{576}}$$

Qisqartirish.

$$x+\frac{25}{24}=\frac{\sqrt{2785}}{24} x+\frac{25}{24}=-\frac{\sqrt{2785}}{24}$$

Tenglamaning ikkala tarafidan \frac{25}{24} ni ayirish.

$$x=\frac{\sqrt{2785}-25}{24} x=\frac{-\sqrt{2785}-25}{24}$$