c^2-10c-125=0 eching

$${ c }^{ 2 } -10c-125 = 0$$

c uchun yechish

$$c=5\sqrt{6}+5\approx 17,247448714$$

$$c=5-5\sqrt{6}\approx -7,247448714$$

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.

$$c^{2}-10c-125=0$$

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 -125 ni c bilan almashtiring.

$$c=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-125\right)}}{2}$$

-10 kvadratini chiqarish.

$$c=\frac{-\left(-10\right)±\sqrt{100-4\left(-125\right)}}{2}$$

-4 ni -125 marotabaga ko'paytirish.

$$c=\frac{-\left(-10\right)±\sqrt{100+500}}{2}$$

100 ni 500 ga qo'shish.

$$c=\frac{-\left(-10\right)±\sqrt{600}}{2}$$

600 ning kvadrat ildizini chiqarish.

$$c=\frac{-\left(-10\right)±10\sqrt{6}}{2}$$

-10 ning teskarisi 10 ga teng.

$$c=\frac{10±10\sqrt{6}}{2}$$

c=\frac{10±10\sqrt{6}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 10\sqrt{6} ga qo'shish.

$$c=\frac{10\sqrt{6}+10}{2}$$

10+10\sqrt{6} ni 2 ga bo'lish.

$$c=5\sqrt{6}+5$$

c=\frac{10±10\sqrt{6}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 10\sqrt{6} ni ayirish.

$$c=\frac{10-10\sqrt{6}}{2}$$

10-10\sqrt{6} ni 2 ga bo'lish.

$$c=5-5\sqrt{6}$$

Tenglama yechildi.

$$c=5\sqrt{6}+5 c=5-5\sqrt{6}$$

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.

$$c^{2}-10c-125=0$$

125 ni tenglamaning ikkala tarafiga qo'shish.

$$c^{2}-10c-125-\left(-125\right)=-\left(-125\right)$$

O‘zidan -125 ayirilsa 0 qoladi.

$$c^{2}-10c=-\left(-125\right)$$

0 dan -125 ni ayirish.

$$c^{2}-10c=125$$

-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.

$$c^{2}-10c+\left(-5\right)^{2}=125+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$c^{2}-10c+25=125+25$$

125 ni 25 ga qo'shish.

$$c^{2}-10c+25=150$$

c^{2}-10c+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.

$$\left(c-5\right)^{2}=150$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(c-5\right)^{2}}=\sqrt{150}$$

Qisqartirish.

$$c-5=5\sqrt{6} c-5=-5\sqrt{6}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$c=5\sqrt{6}+5 c=5-5\sqrt{6}$$