54(1+x)(1+x)=1215 eching

$$54(1+x)(1+x)=1215$$

x uchun yechish

$$x=\frac{3\sqrt{10}}{2}-1\approx 3,74341649$$

$$x=-\frac{3\sqrt{10}}{2}-1\approx -5,74341649$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(1+x\right)^{2} hosil qilish uchun 1+x va 1+x ni ko'paytirish.

$$54\left(1+x\right)^{2}=1215$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+x\right)^{2} kengaytirilishi uchun ishlating.

$$54\left(1+2x+x^{2}\right)=1215$$

54 ga 1+2x+x^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$54+108x+54x^{2}=1215$$

Ikkala tarafdan 1215 ni ayirish.

$$54+108x+54x^{2}-1215=0$$

-1161 olish uchun 54 dan 1215 ni ayirish.

$$-1161+108x+54x^{2}=0$$

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.

$$54x^{2}+108x-1161=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 54 ni a, 108 ni b va -1161 ni c bilan almashtiring.

$$x=\frac{-108±\sqrt{108^{2}-4\times 54\left(-1161\right)}}{2\times 54}$$

108 kvadratini chiqarish.

$$x=\frac{-108±\sqrt{11664-4\times 54\left(-1161\right)}}{2\times 54}$$

-4 ni 54 marotabaga ko'paytirish.

$$x=\frac{-108±\sqrt{11664-216\left(-1161\right)}}{2\times 54}$$

-216 ni -1161 marotabaga ko'paytirish.

$$x=\frac{-108±\sqrt{11664+250776}}{2\times 54}$$

11664 ni 250776 ga qo'shish.

$$x=\frac{-108±\sqrt{262440}}{2\times 54}$$

262440 ning kvadrat ildizini chiqarish.

$$x=\frac{-108±162\sqrt{10}}{2\times 54}$$

2 ni 54 marotabaga ko'paytirish.

$$x=\frac{-108±162\sqrt{10}}{108}$$

x=\frac{-108±162\sqrt{10}}{108} tenglamasini yeching, bunda ± musbat. -108 ni 162\sqrt{10} ga qo'shish.

$$x=\frac{162\sqrt{10}-108}{108}$$

-108+162\sqrt{10} ni 108 ga bo'lish.

$$x=\frac{3\sqrt{10}}{2}-1$$

x=\frac{-108±162\sqrt{10}}{108} tenglamasini yeching, bunda ± manfiy. -108 dan 162\sqrt{10} ni ayirish.

$$x=\frac{-162\sqrt{10}-108}{108}$$

-108-162\sqrt{10} ni 108 ga bo'lish.

$$x=-\frac{3\sqrt{10}}{2}-1$$

Tenglama yechildi.

$$x=\frac{3\sqrt{10}}{2}-1 x=-\frac{3\sqrt{10}}{2}-1$$

Kvadratni yakunlash bo'yicha qadamlar

\left(1+x\right)^{2} hosil qilish uchun 1+x va 1+x ni ko'paytirish.

$$54\left(1+x\right)^{2}=1215$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+x\right)^{2} kengaytirilishi uchun ishlating.

$$54\left(1+2x+x^{2}\right)=1215$$

54 ga 1+2x+x^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$54+108x+54x^{2}=1215$$

Ikkala tarafdan 54 ni ayirish.

$$108x+54x^{2}=1215-54$$

1161 olish uchun 1215 dan 54 ni ayirish.

$$108x+54x^{2}=1161$$

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

$$54x^{2}+108x=1161$$

Ikki tarafini 54 ga bo‘ling.

$$\frac{54x^{2}+108x}{54}=\frac{1161}{54}$$

54 ga bo'lish 54 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{108}{54}x=\frac{1161}{54}$$

108 ni 54 ga bo'lish.

$$x^{2}+2x=\frac{1161}{54}$$

\frac{1161}{54} ulushini 27 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+2x=\frac{43}{2}$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=\frac{43}{2}+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=\frac{43}{2}+1$$

\frac{43}{2} ni 1 ga qo'shish.

$$x^{2}+2x+1=\frac{45}{2}$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=\frac{45}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{45}{2}}$$

Qisqartirish.

$$x+1=\frac{3\sqrt{10}}{2} x+1=-\frac{3\sqrt{10}}{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{3\sqrt{10}}{2}-1 x=-\frac{3\sqrt{10}}{2}-1$$