x^2-8x+9=0 eching

$$x ^ { 2 } - 8 x + 9 = 0$$

x uchun yechish

$$x=\sqrt{7}+4\approx 6,645751311$$

$$x=4-\sqrt{7}\approx 1,354248689$$

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.

$$x^{2}-8x+9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -8 ni b va 9 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 9}}{2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 9}}{2}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-36}}{2}$$

64 ni -36 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±2\sqrt{7}}{2}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±2\sqrt{7}}{2}$$

x=\frac{8±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. 8 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}+8}{2}$$

8+2\sqrt{7} ni 2 ga bo'lish.

$$x=\sqrt{7}+4$$

x=\frac{8±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. 8 dan 2\sqrt{7} ni ayirish.

$$x=\frac{8-2\sqrt{7}}{2}$$

8-2\sqrt{7} ni 2 ga bo'lish.

$$x=4-\sqrt{7}$$

Tenglama yechildi.

$$x=\sqrt{7}+4 x=4-\sqrt{7}$$

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.

$$x^{2}-8x+9=0$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$x^{2}-8x+9-9=-9$$

O‘zidan 9 ayirilsa 0 qoladi.

$$x^{2}-8x=-9$$

-8 ni bo‘lish, x shartining koeffitsienti, 2 ga -4 olish uchun. Keyin, -4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-8x+\left(-4\right)^{2}=-9+\left(-4\right)^{2}$$

-4 kvadratini chiqarish.

$$x^{2}-8x+16=-9+16$$

-9 ni 16 ga qo'shish.

$$x^{2}-8x+16=7$$

x^{2}-8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-4\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$x-4=\sqrt{7} x-4=-\sqrt{7}$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{7}+4 x=4-\sqrt{7}$$