d^2-10d+5=0 eching

$$d ^ { 2 } - 10 d + 5 = 0$$

d uchun yechish

$$d=2\sqrt{5}+5\approx 9,472135955$$

$$d=5-2\sqrt{5}\approx 0,527864045$$

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.

$$d^{2}-10d+5=0$$

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 5 ni c bilan almashtiring.

$$d=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5}}{2}$$

-10 kvadratini chiqarish.

$$d=\frac{-\left(-10\right)±\sqrt{100-4\times 5}}{2}$$

-4 ni 5 marotabaga ko'paytirish.

$$d=\frac{-\left(-10\right)±\sqrt{100-20}}{2}$$

100 ni -20 ga qo'shish.

$$d=\frac{-\left(-10\right)±\sqrt{80}}{2}$$

80 ning kvadrat ildizini chiqarish.

$$d=\frac{-\left(-10\right)±4\sqrt{5}}{2}$$

-10 ning teskarisi 10 ga teng.

$$d=\frac{10±4\sqrt{5}}{2}$$

d=\frac{10±4\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 4\sqrt{5} ga qo'shish.

$$d=\frac{4\sqrt{5}+10}{2}$$

10+4\sqrt{5} ni 2 ga bo'lish.

$$d=2\sqrt{5}+5$$

d=\frac{10±4\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 4\sqrt{5} ni ayirish.

$$d=\frac{10-4\sqrt{5}}{2}$$

10-4\sqrt{5} ni 2 ga bo'lish.

$$d=5-2\sqrt{5}$$

Tenglama yechildi.

$$d=2\sqrt{5}+5 d=5-2\sqrt{5}$$

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.

$$d^{2}-10d+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$d^{2}-10d+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$d^{2}-10d=-5$$

-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.

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

-5 kvadratini chiqarish.

$$d^{2}-10d+25=-5+25$$

-5 ni 25 ga qo'shish.

$$d^{2}-10d+25=20$$

d^{2}-10d+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.

$$\left(d-5\right)^{2}=20$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(d-5\right)^{2}}=\sqrt{20}$$

Qisqartirish.

$$d-5=2\sqrt{5} d-5=-2\sqrt{5}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$d=2\sqrt{5}+5 d=5-2\sqrt{5}$$