25x^2+30x=12 eching

$$25 { x }^{ 2 } +30x=12$$

x uchun yechish

$$x=\frac{\sqrt{21}-3}{5}\approx 0,316515139$$

$$x=\frac{-\sqrt{21}-3}{5}\approx -1,516515139$$

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.

$$25x^{2}+30x=12$$

Tenglamaning ikkala tarafidan 12 ni ayirish.

$$25x^{2}+30x-12=12-12$$

O‘zidan 12 ayirilsa 0 qoladi.

$$25x^{2}+30x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, 30 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-30±\sqrt{30^{2}-4\times 25\left(-12\right)}}{2\times 25}$$

30 kvadratini chiqarish.

$$x=\frac{-30±\sqrt{900-4\times 25\left(-12\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$x=\frac{-30±\sqrt{900-100\left(-12\right)}}{2\times 25}$$

-100 ni -12 marotabaga ko'paytirish.

$$x=\frac{-30±\sqrt{900+1200}}{2\times 25}$$

900 ni 1200 ga qo'shish.

$$x=\frac{-30±\sqrt{2100}}{2\times 25}$$

2100 ning kvadrat ildizini chiqarish.

$$x=\frac{-30±10\sqrt{21}}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{-30±10\sqrt{21}}{50}$$

x=\frac{-30±10\sqrt{21}}{50} tenglamasini yeching, bunda ± musbat. -30 ni 10\sqrt{21} ga qo'shish.

$$x=\frac{10\sqrt{21}-30}{50}$$

-30+10\sqrt{21} ni 50 ga bo'lish.

$$x=\frac{\sqrt{21}-3}{5}$$

x=\frac{-30±10\sqrt{21}}{50} tenglamasini yeching, bunda ± manfiy. -30 dan 10\sqrt{21} ni ayirish.

$$x=\frac{-10\sqrt{21}-30}{50}$$

-30-10\sqrt{21} ni 50 ga bo'lish.

$$x=\frac{-\sqrt{21}-3}{5}$$

Tenglama yechildi.

$$x=\frac{\sqrt{21}-3}{5} x=\frac{-\sqrt{21}-3}{5}$$

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.

$$25x^{2}+30x=12$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}+30x}{25}=\frac{12}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{30}{25}x=\frac{12}{25}$$

\frac{30}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{6}{5}x=\frac{12}{25}$$

\frac{6}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{5} olish uchun. Keyin, \frac{3}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{12}{25}+\left(\frac{3}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{5} kvadratini chiqarish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{12+9}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{12}{25} ni \frac{9}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{21}{25}$$

x^{2}+\frac{6}{5}x+\frac{9}{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.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{21}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{21}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{\sqrt{21}}{5} x+\frac{3}{5}=-\frac{\sqrt{21}}{5}$$

Tenglamaning ikkala tarafidan \frac{3}{5} ni ayirish.

$$x=\frac{\sqrt{21}-3}{5} x=\frac{-\sqrt{21}-3}{5}$$