4x^2+7x=1 eching

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

x uchun yechish

$$x=\frac{\sqrt{65}-7}{8}\approx 0,132782219$$

$$x=\frac{-\sqrt{65}-7}{8}\approx -1,882782219$$

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.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 7 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-7±\sqrt{7^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

7 kvadratini chiqarish.

$$x=\frac{-7±\sqrt{49-4\times 4\left(-1\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-16\left(-1\right)}}{2\times 4}$$

-16 ni -1 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49+16}}{2\times 4}$$

49 ni 16 ga qo'shish.

$$x=\frac{-7±\sqrt{65}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{65}}{8}$$

x=\frac{-7±\sqrt{65}}{8} tenglamasini yeching, bunda ± musbat. -7 ni \sqrt{65} ga qo'shish.

$$x=\frac{\sqrt{65}-7}{8}$$

x=\frac{-7±\sqrt{65}}{8} tenglamasini yeching, bunda ± manfiy. -7 dan \sqrt{65} ni ayirish.

$$x=\frac{-\sqrt{65}-7}{8}$$

Tenglama yechildi.

$$x=\frac{\sqrt{65}-7}{8} x=\frac{-\sqrt{65}-7}{8}$$

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.

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

Ikki tarafini 4 ga bo‘ling.

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

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

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

\frac{7}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{8} olish uchun. Keyin, \frac{7}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{7}{4}x+\left(\frac{7}{8}\right)^{2}=\frac{1}{4}+\left(\frac{7}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{8} kvadratini chiqarish.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{1}{4}+\frac{49}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{4} ni \frac{49}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{65}{64}$$

x^{2}+\frac{7}{4}x+\frac{49}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{7}{8}\right)^{2}=\frac{65}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{8}\right)^{2}}=\sqrt{\frac{65}{64}}$$

Qisqartirish.

$$x+\frac{7}{8}=\frac{\sqrt{65}}{8} x+\frac{7}{8}=-\frac{\sqrt{65}}{8}$$

Tenglamaning ikkala tarafidan \frac{7}{8} ni ayirish.

$$x=\frac{\sqrt{65}-7}{8} x=\frac{-\sqrt{65}-7}{8}$$