1+x^2=5x eching

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

x uchun yechish

$$x = \frac{\sqrt{21} + 5}{2} \approx 4,791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0,208712153$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 5x ni ayirish.

$$1+x^{2}-5x=0$$

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.

$$x^{2}-5x+1=0$$

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 1 ni c bilan almashtiring.

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

-5 kvadratini chiqarish.

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

25 ni -4 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{21}}{2}$$

x=\frac{5±\sqrt{21}}{2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{21} ga qo'shish.

$$x=\frac{\sqrt{21}+5}{2}$$

x=\frac{5±\sqrt{21}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{21} ni ayirish.

$$x=\frac{5-\sqrt{21}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 5x ni ayirish.

$$1+x^{2}-5x=0$$

Ikkala tarafdan 1 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$x^{2}-5x=-1$$

-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.

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

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

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

-1 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

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.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$