22x^2+24x-9=0 eching

$$22 x ^ { 2 } + 24 x - 9 = 0$$

x uchun yechish

$$x=\frac{3\sqrt{38}}{22}-\frac{6}{11}\approx 0,295147364$$

$$x=-\frac{3\sqrt{38}}{22}-\frac{6}{11}\approx -1,386056455$$

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.

$$22x^{2}+24x-9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 22 ni a, 24 ni b va -9 ni c bilan almashtiring.

$$x=\frac{-24±\sqrt{24^{2}-4\times 22\left(-9\right)}}{2\times 22}$$

24 kvadratini chiqarish.

$$x=\frac{-24±\sqrt{576-4\times 22\left(-9\right)}}{2\times 22}$$

-4 ni 22 marotabaga ko'paytirish.

$$x=\frac{-24±\sqrt{576-88\left(-9\right)}}{2\times 22}$$

-88 ni -9 marotabaga ko'paytirish.

$$x=\frac{-24±\sqrt{576+792}}{2\times 22}$$

576 ni 792 ga qo'shish.

$$x=\frac{-24±\sqrt{1368}}{2\times 22}$$

1368 ning kvadrat ildizini chiqarish.

$$x=\frac{-24±6\sqrt{38}}{2\times 22}$$

2 ni 22 marotabaga ko'paytirish.

$$x=\frac{-24±6\sqrt{38}}{44}$$

x=\frac{-24±6\sqrt{38}}{44} tenglamasini yeching, bunda ± musbat. -24 ni 6\sqrt{38} ga qo'shish.

$$x=\frac{6\sqrt{38}-24}{44}$$

-24+6\sqrt{38} ni 44 ga bo'lish.

$$x=\frac{3\sqrt{38}}{22}-\frac{6}{11}$$

x=\frac{-24±6\sqrt{38}}{44} tenglamasini yeching, bunda ± manfiy. -24 dan 6\sqrt{38} ni ayirish.

$$x=\frac{-6\sqrt{38}-24}{44}$$

-24-6\sqrt{38} ni 44 ga bo'lish.

$$x=-\frac{3\sqrt{38}}{22}-\frac{6}{11}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{38}}{22}-\frac{6}{11} x=-\frac{3\sqrt{38}}{22}-\frac{6}{11}$$

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.

$$22x^{2}+24x-9=0$$

9 ni tenglamaning ikkala tarafiga qo'shish.

$$22x^{2}+24x-9-\left(-9\right)=-\left(-9\right)$$

O‘zidan -9 ayirilsa 0 qoladi.

$$22x^{2}+24x=-\left(-9\right)$$

0 dan -9 ni ayirish.

$$22x^{2}+24x=9$$

Ikki tarafini 22 ga bo‘ling.

$$\frac{22x^{2}+24x}{22}=\frac{9}{22}$$

22 ga bo'lish 22 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{24}{22}x=\frac{9}{22}$$

\frac{24}{22} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{12}{11}x=\frac{9}{22}$$

\frac{12}{11} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{6}{11} olish uchun. Keyin, \frac{6}{11} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{12}{11}x+\left(\frac{6}{11}\right)^{2}=\frac{9}{22}+\left(\frac{6}{11}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{6}{11} kvadratini chiqarish.

$$x^{2}+\frac{12}{11}x+\frac{36}{121}=\frac{9}{22}+\frac{36}{121}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{9}{22} ni \frac{36}{121} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{12}{11}x+\frac{36}{121}=\frac{171}{242}$$

x^{2}+\frac{12}{11}x+\frac{36}{121} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{6}{11}\right)^{2}=\frac{171}{242}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{6}{11}\right)^{2}}=\sqrt{\frac{171}{242}}$$

Qisqartirish.

$$x+\frac{6}{11}=\frac{3\sqrt{38}}{22} x+\frac{6}{11}=-\frac{3\sqrt{38}}{22}$$

Tenglamaning ikkala tarafidan \frac{6}{11} ni ayirish.

$$x=\frac{3\sqrt{38}}{22}-\frac{6}{11} x=-\frac{3\sqrt{38}}{22}-\frac{6}{11}$$