x^2-x+4=8 eching

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

x uchun yechish

$$x = \frac{\sqrt{17} + 1}{2} \approx 2,561552813$$

$$x=\frac{1-\sqrt{17}}{2}\approx -1,561552813$$

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.

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

Tenglamaning ikkala tarafidan 8 ni ayirish.

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

O‘zidan 8 ayirilsa 0 qoladi.

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

4 dan 8 ni ayirish.

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

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

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

-4 ni -4 marotabaga ko'paytirish.

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

1 ni 16 ga qo'shish.

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

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{17}}{2}$$

x=\frac{1±\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. 1 ni \sqrt{17} ga qo'shish.

$$x=\frac{\sqrt{17}+1}{2}$$

x=\frac{1±\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. 1 dan \sqrt{17} ni ayirish.

$$x=\frac{1-\sqrt{17}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{17}+1}{2} x=\frac{1-\sqrt{17}}{2}$$

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.

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

Tenglamaning ikkala tarafidan 4 ni ayirish.

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

O‘zidan 4 ayirilsa 0 qoladi.

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

8 dan 4 ni ayirish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

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

4 ni \frac{1}{4} ga qo'shish.

$$x^{2}-x+\frac{1}{4}=\frac{17}{4}$$

x^{2}-x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{17}}{2} x-\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{17}+1}{2} x=\frac{1-\sqrt{17}}{2}$$