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

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

x uchun yechish

$$x=\sqrt{3}+\frac{3}{2}\approx 3,232050808$$

$$x=\frac{3}{2}-\sqrt{3}\approx -0,232050808$$

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.

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

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

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

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+16\times 3}}{2\left(-4\right)}$$

16 ni 3 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+48}}{2\left(-4\right)}$$

144 ni 48 ga qo'shish.

$$x=\frac{-12±\sqrt{192}}{2\left(-4\right)}$$

192 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±8\sqrt{3}}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$x=\frac{-12±8\sqrt{3}}{-8}$$

x=\frac{-12±8\sqrt{3}}{-8} tenglamasini yeching, bunda ± musbat. -12 ni 8\sqrt{3} ga qo'shish.

$$x=\frac{8\sqrt{3}-12}{-8}$$

-12+8\sqrt{3} ni -8 ga bo'lish.

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

x=\frac{-12±8\sqrt{3}}{-8} tenglamasini yeching, bunda ± manfiy. -12 dan 8\sqrt{3} ni ayirish.

$$x=\frac{-8\sqrt{3}-12}{-8}$$

-12-8\sqrt{3} ni -8 ga bo'lish.

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

Tenglama yechildi.

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

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.

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

Tenglamaning ikkala tarafidan 3 ni ayirish.

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

O‘zidan 3 ayirilsa 0 qoladi.

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

Ikki tarafini -4 ga bo‘ling.

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

-4 ga bo'lish -4 ga ko'paytirishni bekor qiladi.

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

12 ni -4 ga bo'lish.

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

-3 ni -4 ga bo'lish.

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

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

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

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

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

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

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