2222x^2+12x-23=0 eching

$$2222 { x }^{ 2 } +12x-23=0$$

x uchun yechish

$$x=\frac{\sqrt{51142}}{2222}-\frac{3}{1111}\approx 0,099075594$$

$$x=-\frac{\sqrt{51142}}{2222}-\frac{3}{1111}\approx -0,104476134$$

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.

$$2222x^{2}+12x-23=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2222 ni a, 12 ni b va -23 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2222\left(-23\right)}}{2\times 2222}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 2222\left(-23\right)}}{2\times 2222}$$

-4 ni 2222 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-8888\left(-23\right)}}{2\times 2222}$$

-8888 ni -23 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+204424}}{2\times 2222}$$

144 ni 204424 ga qo'shish.

$$x=\frac{-12±\sqrt{204568}}{2\times 2222}$$

204568 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±2\sqrt{51142}}{2\times 2222}$$

2 ni 2222 marotabaga ko'paytirish.

$$x=\frac{-12±2\sqrt{51142}}{4444}$$

x=\frac{-12±2\sqrt{51142}}{4444} tenglamasini yeching, bunda ± musbat. -12 ni 2\sqrt{51142} ga qo'shish.

$$x=\frac{2\sqrt{51142}-12}{4444}$$

-12+2\sqrt{51142} ni 4444 ga bo'lish.

$$x=\frac{\sqrt{51142}}{2222}-\frac{3}{1111}$$

x=\frac{-12±2\sqrt{51142}}{4444} tenglamasini yeching, bunda ± manfiy. -12 dan 2\sqrt{51142} ni ayirish.

$$x=\frac{-2\sqrt{51142}-12}{4444}$$

-12-2\sqrt{51142} ni 4444 ga bo'lish.

$$x=-\frac{\sqrt{51142}}{2222}-\frac{3}{1111}$$

Tenglama yechildi.

$$x=\frac{\sqrt{51142}}{2222}-\frac{3}{1111} x=-\frac{\sqrt{51142}}{2222}-\frac{3}{1111}$$

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.

$$2222x^{2}+12x-23=0$$

23 ni tenglamaning ikkala tarafiga qo'shish.

$$2222x^{2}+12x-23-\left(-23\right)=-\left(-23\right)$$

O‘zidan -23 ayirilsa 0 qoladi.

$$2222x^{2}+12x=-\left(-23\right)$$

0 dan -23 ni ayirish.

$$2222x^{2}+12x=23$$

Ikki tarafini 2222 ga bo‘ling.

$$\frac{2222x^{2}+12x}{2222}=\frac{23}{2222}$$

2222 ga bo'lish 2222 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{12}{2222}x=\frac{23}{2222}$$

\frac{12}{2222} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{6}{1111}x=\frac{23}{2222}$$

\frac{6}{1111} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{1111} olish uchun. Keyin, \frac{3}{1111} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{6}{1111}x+\left(\frac{3}{1111}\right)^{2}=\frac{23}{2222}+\left(\frac{3}{1111}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{1111} kvadratini chiqarish.

$$x^{2}+\frac{6}{1111}x+\frac{9}{1234321}=\frac{23}{2222}+\frac{9}{1234321}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{23}{2222} ni \frac{9}{1234321} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{6}{1111}x+\frac{9}{1234321}=\frac{25571}{2468642}$$

x^{2}+\frac{6}{1111}x+\frac{9}{1234321} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{1111}\right)^{2}=\frac{25571}{2468642}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{1111}\right)^{2}}=\sqrt{\frac{25571}{2468642}}$$

Qisqartirish.

$$x+\frac{3}{1111}=\frac{\sqrt{51142}}{2222} x+\frac{3}{1111}=-\frac{\sqrt{51142}}{2222}$$

Tenglamaning ikkala tarafidan \frac{3}{1111} ni ayirish.

$$x=\frac{\sqrt{51142}}{2222}-\frac{3}{1111} x=-\frac{\sqrt{51142}}{2222}-\frac{3}{1111}$$