4x^2+x-2=0 eching

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

x uchun yechish

$$x=\frac{\sqrt{33}-1}{8}\approx 0,593070331$$

$$x=\frac{-\sqrt{33}-1}{8}\approx -0,843070331$$

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 1 ni b va -2 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-2\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-16\left(-2\right)}}{2\times 4}$$

-16 ni -2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+32}}{2\times 4}$$

1 ni 32 ga qo'shish.

$$x=\frac{-1±\sqrt{33}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{33}}{8}$$

x=\frac{-1±\sqrt{33}}{8} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{33} ga qo'shish.

$$x=\frac{\sqrt{33}-1}{8}$$

x=\frac{-1±\sqrt{33}}{8} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{33} ni ayirish.

$$x=\frac{-\sqrt{33}-1}{8}$$

Tenglama yechildi.

$$x=\frac{\sqrt{33}-1}{8} x=\frac{-\sqrt{33}-1}{8}$$

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.

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

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

0 dan -2 ni ayirish.

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

Ikki tarafini 4 ga bo‘ling.

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

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

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

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

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

\frac{1}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{8} olish uchun. Keyin, \frac{1}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{8} kvadratini chiqarish.

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

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

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{33}{64}$$

x^{2}+\frac{1}{4}x+\frac{1}{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.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{33}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Qisqartirish.

$$x+\frac{1}{8}=\frac{\sqrt{33}}{8} x+\frac{1}{8}=-\frac{\sqrt{33}}{8}$$

Tenglamaning ikkala tarafidan \frac{1}{8} ni ayirish.

$$x=\frac{\sqrt{33}-1}{8} x=\frac{-\sqrt{33}-1}{8}$$