x^2-8x-1029=0= eching

$${ x }^{ 2 } -8x-1029=0=$$

x uchun yechish

$$x=\sqrt{1045}+4\approx 36,32645975$$

$$x=4-\sqrt{1045}\approx -28,32645975$$

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.

$$x^{2}-8x-1029=0$$

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 -1029 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1029\right)}}{2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1029\right)}}{2}$$

-4 ni -1029 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64+4116}}{2}$$

64 ni 4116 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{4180}}{2}$$

4180 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±2\sqrt{1045}}{2}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±2\sqrt{1045}}{2}$$

x=\frac{8±2\sqrt{1045}}{2} tenglamasini yeching, bunda ± musbat. 8 ni 2\sqrt{1045} ga qo'shish.

$$x=\frac{2\sqrt{1045}+8}{2}$$

8+2\sqrt{1045} ni 2 ga bo'lish.

$$x=\sqrt{1045}+4$$

x=\frac{8±2\sqrt{1045}}{2} tenglamasini yeching, bunda ± manfiy. 8 dan 2\sqrt{1045} ni ayirish.

$$x=\frac{8-2\sqrt{1045}}{2}$$

8-2\sqrt{1045} ni 2 ga bo'lish.

$$x=4-\sqrt{1045}$$

Tenglama yechildi.

$$x=\sqrt{1045}+4 x=4-\sqrt{1045}$$

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.

$$x^{2}-8x-1029=0$$

1029 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-8x-1029-\left(-1029\right)=-\left(-1029\right)$$

O‘zidan -1029 ayirilsa 0 qoladi.

$$x^{2}-8x=-\left(-1029\right)$$

0 dan -1029 ni ayirish.

$$x^{2}-8x=1029$$

-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.

$$x^{2}-8x+\left(-4\right)^{2}=1029+\left(-4\right)^{2}$$

-4 kvadratini chiqarish.

$$x^{2}-8x+16=1029+16$$

1029 ni 16 ga qo'shish.

$$x^{2}-8x+16=1045$$

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.

$$\left(x-4\right)^{2}=1045$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{1045}$$

Qisqartirish.

$$x-4=\sqrt{1045} x-4=-\sqrt{1045}$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{1045}+4 x=4-\sqrt{1045}$$