(5-d)(5+10d)=(5+2d)^2 eching

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

d uchun yechish

$$d = \frac{25}{14} = 1\frac{11}{14} \approx 1,785714286$$

$$d=0$$

Faktorlash qadamlari

5-d ga 5+10d ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5+2d\right)^{2} kengaytirilishi uchun ishlating.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Ikkala tarafdan 25 ni ayirish.

$$25+45d-10d^{2}-25=20d+4d^{2}$$

0 olish uchun 25 dan 25 ni ayirish.

$$45d-10d^{2}=20d+4d^{2}$$

Ikkala tarafdan 20d ni ayirish.

$$45d-10d^{2}-20d=4d^{2}$$

25d ni olish uchun 45d va -20d ni birlashtirish.

$$25d-10d^{2}=4d^{2}$$

Ikkala tarafdan 4d^{2} ni ayirish.

$$25d-10d^{2}-4d^{2}=0$$

-14d^{2} ni olish uchun -10d^{2} va -4d^{2} ni birlashtirish.

$$25d-14d^{2}=0$$

d omili.

$$d\left(25-14d\right)=0$$

Tenglamani yechish uchun d=0 va 25-14d=0 ni yeching.

$$d=0 d=\frac{25}{14}$$

Kvadrat tenglamalaridan foydalanish qoidalari

5-d ga 5+10d ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5+2d\right)^{2} kengaytirilishi uchun ishlating.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Ikkala tarafdan 25 ni ayirish.

$$25+45d-10d^{2}-25=20d+4d^{2}$$

0 olish uchun 25 dan 25 ni ayirish.

$$45d-10d^{2}=20d+4d^{2}$$

Ikkala tarafdan 20d ni ayirish.

$$45d-10d^{2}-20d=4d^{2}$$

25d ni olish uchun 45d va -20d ni birlashtirish.

$$25d-10d^{2}=4d^{2}$$

Ikkala tarafdan 4d^{2} ni ayirish.

$$25d-10d^{2}-4d^{2}=0$$

-14d^{2} ni olish uchun -10d^{2} va -4d^{2} ni birlashtirish.

$$25d-14d^{2}=0$$

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.

$$-14d^{2}+25d=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -14 ni a, 25 ni b va 0 ni c bilan almashtiring.

$$d=\frac{-25±\sqrt{25^{2}}}{2\left(-14\right)}$$

25^{2} ning kvadrat ildizini chiqarish.

$$d=\frac{-25±25}{2\left(-14\right)}$$

2 ni -14 marotabaga ko'paytirish.

$$d=\frac{-25±25}{-28}$$

d=\frac{-25±25}{-28} tenglamasini yeching, bunda ± musbat. -25 ni 25 ga qo'shish.

$$d=\frac{0}{-28}$$

0 ni -28 ga bo'lish.

$$d=0$$

d=\frac{-25±25}{-28} tenglamasini yeching, bunda ± manfiy. -25 dan 25 ni ayirish.

$$d=-\frac{50}{-28}$$

\frac{-50}{-28} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$d=\frac{25}{14}$$

Tenglama yechildi.

$$d=0 d=\frac{25}{14}$$

Kvadratni yakunlash bo'yicha qadamlar

5-d ga 5+10d ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5+2d\right)^{2} kengaytirilishi uchun ishlating.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Ikkala tarafdan 20d ni ayirish.

$$25+45d-10d^{2}-20d=25+4d^{2}$$

25d ni olish uchun 45d va -20d ni birlashtirish.

$$25+25d-10d^{2}=25+4d^{2}$$

Ikkala tarafdan 4d^{2} ni ayirish.

$$25+25d-10d^{2}-4d^{2}=25$$

-14d^{2} ni olish uchun -10d^{2} va -4d^{2} ni birlashtirish.

$$25+25d-14d^{2}=25$$

Ikkala tarafdan 25 ni ayirish.

$$25d-14d^{2}=25-25$$

0 olish uchun 25 dan 25 ni ayirish.

$$25d-14d^{2}=0$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-14d^{2}+25d=0$$

Ikki tarafini -14 ga bo‘ling.

$$\frac{-14d^{2}+25d}{-14}=\frac{0}{-14}$$

-14 ga bo'lish -14 ga ko'paytirishni bekor qiladi.

$$d^{2}+\frac{25}{-14}d=\frac{0}{-14}$$

25 ni -14 ga bo'lish.

$$d^{2}-\frac{25}{14}d=\frac{0}{-14}$$

0 ni -14 ga bo'lish.

$$d^{2}-\frac{25}{14}d=0$$

-\frac{25}{14} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{25}{28} olish uchun. Keyin, -\frac{25}{28} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$d^{2}-\frac{25}{14}d+\left(-\frac{25}{28}\right)^{2}=\left(-\frac{25}{28}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{25}{28} kvadratini chiqarish.

$$d^{2}-\frac{25}{14}d+\frac{625}{784}=\frac{625}{784}$$

d^{2}-\frac{25}{14}d+\frac{625}{784} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(d-\frac{25}{28}\right)^{2}=\frac{625}{784}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(d-\frac{25}{28}\right)^{2}}=\sqrt{\frac{625}{784}}$$

Qisqartirish.

$$d-\frac{25}{28}=\frac{25}{28} d-\frac{25}{28}=-\frac{25}{28}$$

\frac{25}{28} ni tenglamaning ikkala tarafiga qo'shish.

$$d=\frac{25}{14} d=0$$