3x^2+35x+1=63 eching

$$3 { x }^{ 2 } +35x+1=63$$

x uchun yechish

$$x = \frac{\sqrt{1969} - 35}{6} \approx 1,562235911$$

$$x=\frac{-\sqrt{1969}-35}{6}\approx -13,228902577$$

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.

$$3x^{2}+35x+1=63$$

Tenglamaning ikkala tarafidan 63 ni ayirish.

$$3x^{2}+35x+1-63=63-63$$

O‘zidan 63 ayirilsa 0 qoladi.

$$3x^{2}+35x+1-63=0$$

1 dan 63 ni ayirish.

$$3x^{2}+35x-62=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 35 ni b va -62 ni c bilan almashtiring.

$$x=\frac{-35±\sqrt{35^{2}-4\times 3\left(-62\right)}}{2\times 3}$$

35 kvadratini chiqarish.

$$x=\frac{-35±\sqrt{1225-4\times 3\left(-62\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-35±\sqrt{1225-12\left(-62\right)}}{2\times 3}$$

-12 ni -62 marotabaga ko'paytirish.

$$x=\frac{-35±\sqrt{1225+744}}{2\times 3}$$

1225 ni 744 ga qo'shish.

$$x=\frac{-35±\sqrt{1969}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-35±\sqrt{1969}}{6}$$

x=\frac{-35±\sqrt{1969}}{6} tenglamasini yeching, bunda ± musbat. -35 ni \sqrt{1969} ga qo'shish.

$$x=\frac{\sqrt{1969}-35}{6}$$

x=\frac{-35±\sqrt{1969}}{6} tenglamasini yeching, bunda ± manfiy. -35 dan \sqrt{1969} ni ayirish.

$$x=\frac{-\sqrt{1969}-35}{6}$$

Tenglama yechildi.

$$x=\frac{\sqrt{1969}-35}{6} x=\frac{-\sqrt{1969}-35}{6}$$

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.

$$3x^{2}+35x+1=63$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$3x^{2}+35x+1-1=63-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$3x^{2}+35x=63-1$$

63 dan 1 ni ayirish.

$$3x^{2}+35x=62$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+35x}{3}=\frac{62}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{35}{3}x=\frac{62}{3}$$

\frac{35}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{35}{6} olish uchun. Keyin, \frac{35}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{35}{3}x+\left(\frac{35}{6}\right)^{2}=\frac{62}{3}+\left(\frac{35}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{35}{6} kvadratini chiqarish.

$$x^{2}+\frac{35}{3}x+\frac{1225}{36}=\frac{62}{3}+\frac{1225}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{62}{3} ni \frac{1225}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{35}{3}x+\frac{1225}{36}=\frac{1969}{36}$$

x^{2}+\frac{35}{3}x+\frac{1225}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{35}{6}\right)^{2}=\frac{1969}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{35}{6}\right)^{2}}=\sqrt{\frac{1969}{36}}$$

Qisqartirish.

$$x+\frac{35}{6}=\frac{\sqrt{1969}}{6} x+\frac{35}{6}=-\frac{\sqrt{1969}}{6}$$

Tenglamaning ikkala tarafidan \frac{35}{6} ni ayirish.

$$x=\frac{\sqrt{1969}-35}{6} x=\frac{-\sqrt{1969}-35}{6}$$