1+x+x^2=7/4 eching

$$1 + x + x ^ { 2 } = \frac { 7 } { 4 }$$

x uchun yechish

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1,5$$

$$x=\frac{1}{2}=0,5$$

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.

$$x^{2}+x+1=\frac{7}{4}$$

Tenglamaning ikkala tarafidan \frac{7}{4} ni ayirish.

$$x^{2}+x+1-\frac{7}{4}=\frac{7}{4}-\frac{7}{4}$$

O‘zidan \frac{7}{4} ayirilsa 0 qoladi.

$$x^{2}+x+1-\frac{7}{4}=0$$

1 dan \frac{7}{4} ni ayirish.

$$x^{2}+x-\frac{3}{4}=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 1 ni b va -\frac{3}{4} ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-\frac{3}{4}\right)}}{2}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\left(-\frac{3}{4}\right)}}{2}$$

-4 ni -\frac{3}{4} marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+3}}{2}$$

1 ni 3 ga qo'shish.

$$x=\frac{-1±\sqrt{4}}{2}$$

4 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±2}{2}$$

x=\frac{-1±2}{2} tenglamasini yeching, bunda ± musbat. -1 ni 2 ga qo'shish.

$$x=\frac{1}{2}$$

x=\frac{-1±2}{2} tenglamasini yeching, bunda ± manfiy. -1 dan 2 ni ayirish.

$$x=-\frac{3}{2}$$

Tenglama yechildi.

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

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.

$$x^{2}+x+1=\frac{7}{4}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x^{2}+x+1-1=\frac{7}{4}-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$x^{2}+x=\frac{7}{4}-1$$

\frac{7}{4} dan 1 ni ayirish.

$$x^{2}+x=\frac{3}{4}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$x^{2}+x+\frac{1}{4}=\frac{3+1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+x+\frac{1}{4}=1$$

x^{2}+x+\frac{1}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+\frac{1}{2}=1 x+\frac{1}{2}=-1$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

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