(x^2-5x+3)=8 eching

$$( { x }^{ 2 } -5x+3)=8$$

x uchun yechish

$$x = \frac{3 \sqrt{5} + 5}{2} \approx 5,854101966$$

$$x=\frac{5-3\sqrt{5}}{2}\approx -0,854101966$$

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.

$$x^{2}-5x+3=8$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$x^{2}-5x+3-8=8-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$x^{2}-5x+3-8=0$$

3 dan 8 ni ayirish.

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

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

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

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-5\right)}}{2}$$

-4 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+20}}{2}$$

25 ni 20 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{45}}{2}$$

45 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±3\sqrt{5}}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±3\sqrt{5}}{2}$$

x=\frac{5±3\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 5 ni 3\sqrt{5} ga qo'shish.

$$x=\frac{3\sqrt{5}+5}{2}$$

x=\frac{5±3\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan 3\sqrt{5} ni ayirish.

$$x=\frac{5-3\sqrt{5}}{2}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{5}+5}{2} x=\frac{5-3\sqrt{5}}{2}$$

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.

$$x^{2}-5x+3=8$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x^{2}-5x+3-3=8-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$x^{2}-5x=8-3$$

8 dan 3 ni ayirish.

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

-5 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{2} olish uchun. Keyin, -\frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{2} kvadratini chiqarish.

$$x^{2}-5x+\frac{25}{4}=5+\frac{25}{4}$$

5 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=\frac{45}{4}$$

x^{2}-5x+\frac{25}{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.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{45}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{3\sqrt{5}}{2} x-\frac{5}{2}=-\frac{3\sqrt{5}}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3\sqrt{5}+5}{2} x=\frac{5-3\sqrt{5}}{2}$$