2x^2=x+5 eching

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

x uchun yechish

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x ni ayirish.

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

Ikkala tarafdan 5 ni ayirish.

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

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.

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

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+40}}{2\times 2}$$

1 ni 40 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{41}}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{41}}{2\times 2}$$

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.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan x 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.

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

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

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