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

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

x uchun yechish

$$x = \frac{\sqrt{41} + 3}{8} \approx 1,17539053$$

$$x=\frac{3-\sqrt{41}}{8}\approx -0,42539053$$

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.

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

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

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

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+16\times 2}}{2\left(-4\right)}$$

16 ni 2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+32}}{2\left(-4\right)}$$

9 ni 32 ga qo'shish.

$$x=\frac{-3±\sqrt{41}}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{41}}{-8}$$

x=\frac{-3±\sqrt{41}}{-8} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}-3}{-8}$$

-3+\sqrt{41} ni -8 ga bo'lish.

$$x=\frac{3-\sqrt{41}}{8}$$

x=\frac{-3±\sqrt{41}}{-8} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{41} ni ayirish.

$$x=\frac{-\sqrt{41}-3}{-8}$$

-3-\sqrt{41} ni -8 ga bo'lish.

$$x=\frac{\sqrt{41}+3}{8}$$

Tenglama yechildi.

$$x=\frac{3-\sqrt{41}}{8} x=\frac{\sqrt{41}+3}{8}$$

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.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

Ikki tarafini -4 ga bo‘ling.

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

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

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

3 ni -4 ga bo'lish.

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

\frac{-2}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{3}{4}x=\frac{1}{2}$$

-\frac{3}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{8} olish uchun. Keyin, -\frac{3}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{8} kvadratini chiqarish.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{1}{2}+\frac{9}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{9}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{41}{64}$$

x^{2}-\frac{3}{4}x+\frac{9}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{41}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Qisqartirish.

$$x-\frac{3}{8}=\frac{\sqrt{41}}{8} x-\frac{3}{8}=-\frac{\sqrt{41}}{8}$$

\frac{3}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{41}+3}{8} x=\frac{3-\sqrt{41}}{8}$$