2x^2+x-5=0 eching

$$2 { x }^{ 2 } +x-5=0$$

x uchun yechish

$$x = \frac{\sqrt{41} - 1}{4} \approx 1,350781059$$

$$x=\frac{-\sqrt{41}-1}{4}\approx -1,850781059$$

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.

$$2x^{2}+x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 1 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-5\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

-8 ni -5 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+40}}{2\times 2}$$

1 ni 40 ga qo'shish.

$$x=\frac{-1±\sqrt{41}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{41}}{4}$$

x=\frac{-1±\sqrt{41}}{4} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}-1}{4}$$

x=\frac{-1±\sqrt{41}}{4} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{41} ni ayirish.

$$x=\frac{-\sqrt{41}-1}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}-1}{4} x=\frac{-\sqrt{41}-1}{4}$$

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.

$$2x^{2}+x-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}+x-5-\left(-5\right)=-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$2x^{2}+x=-\left(-5\right)$$

0 dan -5 ni ayirish.

$$2x^{2}+x=5$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+x}{2}=\frac{5}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{1}{2}x=\frac{5}{2}$$

\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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{5}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{5}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{41}{16}$$

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.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{41}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{41}}{4} x+\frac{1}{4}=-\frac{\sqrt{41}}{4}$$

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

$$x=\frac{\sqrt{41}-1}{4} x=\frac{-\sqrt{41}-1}{4}$$