x^2-10x+11=0 eching

$${ x }^{ 2 } -10x+11=0$$

x uchun yechish

$$x=\sqrt{14}+5\approx 8,741657387$$

$$x=5-\sqrt{14}\approx 1,258342613$$

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.

$$x^{2}-10x+11=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -10 ni b va 11 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 11}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 11}}{2}$$

-4 ni 11 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-44}}{2}$$

100 ni -44 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{56}}{2}$$

56 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{14}}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{14}}{2}$$

x=\frac{10±2\sqrt{14}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{14} ga qo'shish.

$$x=\frac{2\sqrt{14}+10}{2}$$

10+2\sqrt{14} ni 2 ga bo'lish.

$$x=\sqrt{14}+5$$

x=\frac{10±2\sqrt{14}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{14} ni ayirish.

$$x=\frac{10-2\sqrt{14}}{2}$$

10-2\sqrt{14} ni 2 ga bo'lish.

$$x=5-\sqrt{14}$$

Tenglama yechildi.

$$x=\sqrt{14}+5 x=5-\sqrt{14}$$

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.

$$x^{2}-10x+11=0$$

Tenglamaning ikkala tarafidan 11 ni ayirish.

$$x^{2}-10x+11-11=-11$$

O‘zidan 11 ayirilsa 0 qoladi.

$$x^{2}-10x=-11$$

-10 ni bo‘lish, x shartining koeffitsienti, 2 ga -5 olish uchun. Keyin, -5 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-10x+\left(-5\right)^{2}=-11+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=-11+25$$

-11 ni 25 ga qo'shish.

$$x^{2}-10x+25=14$$

x^{2}-10x+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-5\right)^{2}=14$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{14}$$

Qisqartirish.

$$x-5=\sqrt{14} x-5=-\sqrt{14}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{14}+5 x=5-\sqrt{14}$$