y^2+5y=625 eching

$$y ^ { 2 } + 5 y = 625$$

y uchun yechish

$$y = \frac{5 \sqrt{101} - 5}{2} \approx 22,624689053$$

$$y=\frac{-5\sqrt{101}-5}{2}\approx -27,624689053$$

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.

$$y^{2}+5y=625$$

Tenglamaning ikkala tarafidan 625 ni ayirish.

$$y^{2}+5y-625=625-625$$

O‘zidan 625 ayirilsa 0 qoladi.

$$y^{2}+5y-625=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 5 ni b va -625 ni c bilan almashtiring.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-625\right)}}{2}$$

5 kvadratini chiqarish.

$$y=\frac{-5±\sqrt{25-4\left(-625\right)}}{2}$$

-4 ni -625 marotabaga ko'paytirish.

$$y=\frac{-5±\sqrt{25+2500}}{2}$$

25 ni 2500 ga qo'shish.

$$y=\frac{-5±\sqrt{2525}}{2}$$

2525 ning kvadrat ildizini chiqarish.

$$y=\frac{-5±5\sqrt{101}}{2}$$

y=\frac{-5±5\sqrt{101}}{2} tenglamasini yeching, bunda ± musbat. -5 ni 5\sqrt{101} ga qo'shish.

$$y=\frac{5\sqrt{101}-5}{2}$$

y=\frac{-5±5\sqrt{101}}{2} tenglamasini yeching, bunda ± manfiy. -5 dan 5\sqrt{101} ni ayirish.

$$y=\frac{-5\sqrt{101}-5}{2}$$

Tenglama yechildi.

$$y=\frac{5\sqrt{101}-5}{2} y=\frac{-5\sqrt{101}-5}{2}$$

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.

$$y^{2}+5y=625$$

5 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{2} olish uchun. Keyin, \frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}+5y+\left(\frac{5}{2}\right)^{2}=625+\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$y^{2}+5y+\frac{25}{4}=625+\frac{25}{4}$$

625 ni \frac{25}{4} ga qo'shish.

$$y^{2}+5y+\frac{25}{4}=\frac{2525}{4}$$

y^{2}+5y+\frac{25}{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.

$$\left(y+\frac{5}{2}\right)^{2}=\frac{2525}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+\frac{5}{2}\right)^{2}}=\sqrt{\frac{2525}{4}}$$

Qisqartirish.

$$y+\frac{5}{2}=\frac{5\sqrt{101}}{2} y+\frac{5}{2}=-\frac{5\sqrt{101}}{2}$$

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

$$y=\frac{5\sqrt{101}-5}{2} y=\frac{-5\sqrt{101}-5}{2}$$