-3x^2+4x+12=0 eching

$$- 3 x ^ { 2 } + 4 x + 12 = 0$$

x uchun yechish

$$x = \frac{2 \sqrt{10} + 2}{3} \approx 2,774851773$$

$$x=\frac{2-2\sqrt{10}}{3}\approx -1,44151844$$

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.

$$-3x^{2}+4x+12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, 4 ni b va 12 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+12\times 12}}{2\left(-3\right)}$$

12 ni 12 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+144}}{2\left(-3\right)}$$

16 ni 144 ga qo'shish.

$$x=\frac{-4±\sqrt{160}}{2\left(-3\right)}$$

160 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{10}}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{-4±4\sqrt{10}}{-6}$$

x=\frac{-4±4\sqrt{10}}{-6} tenglamasini yeching, bunda ± musbat. -4 ni 4\sqrt{10} ga qo'shish.

$$x=\frac{4\sqrt{10}-4}{-6}$$

-4+4\sqrt{10} ni -6 ga bo'lish.

$$x=\frac{2-2\sqrt{10}}{3}$$

x=\frac{-4±4\sqrt{10}}{-6} tenglamasini yeching, bunda ± manfiy. -4 dan 4\sqrt{10} ni ayirish.

$$x=\frac{-4\sqrt{10}-4}{-6}$$

-4-4\sqrt{10} ni -6 ga bo'lish.

$$x=\frac{2\sqrt{10}+2}{3}$$

Tenglama yechildi.

$$x=\frac{2-2\sqrt{10}}{3} x=\frac{2\sqrt{10}+2}{3}$$

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.

$$-3x^{2}+4x+12=0$$

Tenglamaning ikkala tarafidan 12 ni ayirish.

$$-3x^{2}+4x+12-12=-12$$

O‘zidan 12 ayirilsa 0 qoladi.

$$-3x^{2}+4x=-12$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3x^{2}+4x}{-3}=-\frac{12}{-3}$$

-3 ga bo'lish -3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{-3}x=-\frac{12}{-3}$$

4 ni -3 ga bo'lish.

$$x^{2}-\frac{4}{3}x=-\frac{12}{-3}$$

-12 ni -3 ga bo'lish.

$$x^{2}-\frac{4}{3}x=4$$

-\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{2}{3} olish uchun. Keyin, -\frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=4+\left(-\frac{2}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{3} kvadratini chiqarish.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=4+\frac{4}{9}$$

4 ni \frac{4}{9} ga qo'shish.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{40}{9}$$

x^{2}-\frac{4}{3}x+\frac{4}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{40}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{40}{9}}$$

Qisqartirish.

$$x-\frac{2}{3}=\frac{2\sqrt{10}}{3} x-\frac{2}{3}=-\frac{2\sqrt{10}}{3}$$

\frac{2}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{2\sqrt{10}+2}{3} x=\frac{2-2\sqrt{10}}{3}$$