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

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

x uchun yechish

$$x=\frac{\sqrt{2}}{2}-1\approx -0,292893219$$

$$x=-\frac{\sqrt{2}}{2}-1\approx -1,707106781$$

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.

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

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

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\times 2}}{2\times 4}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 4\times 2}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-16\times 2}}{2\times 4}$$

-16 ni 2 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-32}}{2\times 4}$$

64 ni -32 ga qo'shish.

$$x=\frac{-8±\sqrt{32}}{2\times 4}$$

32 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±4\sqrt{2}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-8±4\sqrt{2}}{8}$$

x=\frac{-8±4\sqrt{2}}{8} tenglamasini yeching, bunda ± musbat. -8 ni 4\sqrt{2} ga qo'shish.

$$x=\frac{4\sqrt{2}-8}{8}$$

-8+4\sqrt{2} ni 8 ga bo'lish.

$$x=\frac{\sqrt{2}}{2}-1$$

x=\frac{-8±4\sqrt{2}}{8} tenglamasini yeching, bunda ± manfiy. -8 dan 4\sqrt{2} ni ayirish.

$$x=\frac{-4\sqrt{2}-8}{8}$$

-8-4\sqrt{2} ni 8 ga bo'lish.

$$x=-\frac{\sqrt{2}}{2}-1$$

Tenglama yechildi.

$$x=\frac{\sqrt{2}}{2}-1 x=-\frac{\sqrt{2}}{2}-1$$

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.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

Ikki tarafini 4 ga bo‘ling.

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

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

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

8 ni 4 ga bo'lish.

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

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

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

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

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

1 kvadratini chiqarish.

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

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

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

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

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=\frac{\sqrt{2}}{2} x+1=-\frac{\sqrt{2}}{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{\sqrt{2}}{2}-1 x=-\frac{\sqrt{2}}{2}-1$$