x^2-9x+10=0 eching

$$x ^ { 2 } - 9 x + 10 = 0$$

x uchun yechish

$$x = \frac{\sqrt{41} + 9}{2} \approx 7,701562119$$

$$x = \frac{9 - \sqrt{41}}{2} \approx 1,298437881$$

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.

$$x^{2}-9x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -9 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10}}{2}$$

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 10}}{2}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2}$$

81 ni -40 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±\sqrt{41}}{2}$$

x=\frac{9±\sqrt{41}}{2} tenglamasini yeching, bunda ± musbat. 9 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+9}{2}$$

x=\frac{9±\sqrt{41}}{2} tenglamasini yeching, bunda ± manfiy. 9 dan \sqrt{41} ni ayirish.

$$x=\frac{9-\sqrt{41}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}+9}{2} x=\frac{9-\sqrt{41}}{2}$$

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.

$$x^{2}-9x+10=0$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$x^{2}-9x+10-10=-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$x^{2}-9x=-10$$

-9 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{9}{2} olish uchun. Keyin, -\frac{9}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-10+\left(-\frac{9}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{9}{2} kvadratini chiqarish.

$$x^{2}-9x+\frac{81}{4}=-10+\frac{81}{4}$$

-10 ni \frac{81}{4} ga qo'shish.

$$x^{2}-9x+\frac{81}{4}=\frac{41}{4}$$

x^{2}-9x+\frac{81}{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.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{41}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Qisqartirish.

$$x-\frac{9}{2}=\frac{\sqrt{41}}{2} x-\frac{9}{2}=-\frac{\sqrt{41}}{2}$$

\frac{9}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{41}+9}{2} x=\frac{9-\sqrt{41}}{2}$$