4x^2+2x-8=0 eching

$$4 x ^ { 2 } + 2 x - 8 = 0$$

x uchun yechish

$$x = \frac{\sqrt{33} - 1}{4} \approx 1,186140662$$

$$x=\frac{-\sqrt{33}-1}{4}\approx -1,686140662$$

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.

$$4x^{2}+2x-8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 2 ni b va -8 ni c bilan almashtiring.

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

2 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-16\left(-8\right)}}{2\times 4}$$

-16 ni -8 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+128}}{2\times 4}$$

4 ni 128 ga qo'shish.

$$x=\frac{-2±\sqrt{132}}{2\times 4}$$

132 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{33}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-2±2\sqrt{33}}{8}$$

x=\frac{-2±2\sqrt{33}}{8} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{33} ga qo'shish.

$$x=\frac{2\sqrt{33}-2}{8}$$

-2+2\sqrt{33} ni 8 ga bo'lish.

$$x=\frac{\sqrt{33}-1}{4}$$

x=\frac{-2±2\sqrt{33}}{8} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{33} ni ayirish.

$$x=\frac{-2\sqrt{33}-2}{8}$$

-2-2\sqrt{33} ni 8 ga bo'lish.

$$x=\frac{-\sqrt{33}-1}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$

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.

$$4x^{2}+2x-8=0$$

8 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -8 ayirilsa 0 qoladi.

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

0 dan -8 ni ayirish.

$$4x^{2}+2x=8$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+2x}{4}=\frac{8}{4}$$

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

$$x^{2}+\frac{2}{4}x=\frac{8}{4}$$

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

$$x^{2}+\frac{1}{2}x=\frac{8}{4}$$

8 ni 4 ga bo'lish.

$$x^{2}+\frac{1}{2}x=2$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=2+\frac{1}{16}$$

2 ni \frac{1}{16} ga qo'shish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{33}{16}$$

x^{2}+\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{33}}{4} x+\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$