2a^2-a-2=0 eching

$$2 a ^ { 2 } - a - 2 = 0$$

a uchun yechish

$$a = \frac{\sqrt{17} + 1}{4} \approx 1,280776406$$

$$a=\frac{1-\sqrt{17}}{4}\approx -0,780776406$$

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.

$$2a^{2}-a-2=0$$

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 -2 ni c bilan almashtiring.

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

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni -2 marotabaga ko'paytirish.

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

1 ni 16 ga qo'shish.

$$a=\frac{-\left(-1\right)±\sqrt{17}}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$a=\frac{1±\sqrt{17}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$a=\frac{1±\sqrt{17}}{4}$$

a=\frac{1±\sqrt{17}}{4} tenglamasini yeching, bunda ± musbat. 1 ni \sqrt{17} ga qo'shish.

$$a=\frac{\sqrt{17}+1}{4}$$

a=\frac{1±\sqrt{17}}{4} tenglamasini yeching, bunda ± manfiy. 1 dan \sqrt{17} ni ayirish.

$$a=\frac{1-\sqrt{17}}{4}$$

Tenglama yechildi.

$$a=\frac{\sqrt{17}+1}{4} a=\frac{1-\sqrt{17}}{4}$$

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.

$$2a^{2}-a-2=0$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$2a^{2}-a-2-\left(-2\right)=-\left(-2\right)$$

O‘zidan -2 ayirilsa 0 qoladi.

$$2a^{2}-a=-\left(-2\right)$$

0 dan -2 ni ayirish.

$$2a^{2}-a=2$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2a^{2}-a}{2}=\frac{2}{2}$$

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

$$a^{2}-\frac{1}{2}a=\frac{2}{2}$$

2 ni 2 ga bo'lish.

$$a^{2}-\frac{1}{2}a=1$$

-\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.

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

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

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=1+\frac{1}{16}$$

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

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{17}{16}$$

a^{2}-\frac{1}{2}a+\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.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Qisqartirish.

$$a-\frac{1}{4}=\frac{\sqrt{17}}{4} a-\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

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

$$a=\frac{\sqrt{17}+1}{4} a=\frac{1-\sqrt{17}}{4}$$