6x^2+4x-3=1 eching

$$6 x ^ { 2 } + 4 x - 3 = 1$$

x uchun yechish

$$x=\frac{\sqrt{7}-1}{3}\approx 0,54858377$$

$$x=\frac{-\sqrt{7}-1}{3}\approx -1,215250437$$

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.

$$6x^{2}+4x-3=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$6x^{2}+4x-3-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$6x^{2}+4x-3-1=0$$

-3 dan 1 ni ayirish.

$$6x^{2}+4x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, 4 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 6\left(-4\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-24\left(-4\right)}}{2\times 6}$$

-24 ni -4 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+96}}{2\times 6}$$

16 ni 96 ga qo'shish.

$$x=\frac{-4±\sqrt{112}}{2\times 6}$$

112 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{7}}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{-4±4\sqrt{7}}{12}$$

x=\frac{-4±4\sqrt{7}}{12} tenglamasini yeching, bunda ± musbat. -4 ni 4\sqrt{7} ga qo'shish.

$$x=\frac{4\sqrt{7}-4}{12}$$

-4+4\sqrt{7} ni 12 ga bo'lish.

$$x=\frac{\sqrt{7}-1}{3}$$

x=\frac{-4±4\sqrt{7}}{12} tenglamasini yeching, bunda ± manfiy. -4 dan 4\sqrt{7} ni ayirish.

$$x=\frac{-4\sqrt{7}-4}{12}$$

-4-4\sqrt{7} ni 12 ga bo'lish.

$$x=\frac{-\sqrt{7}-1}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{7}-1}{3} x=\frac{-\sqrt{7}-1}{3}$$

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.

$$6x^{2}+4x-3=1$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$6x^{2}+4x-3-\left(-3\right)=1-\left(-3\right)$$

O‘zidan -3 ayirilsa 0 qoladi.

$$6x^{2}+4x=1-\left(-3\right)$$

1 dan -3 ni ayirish.

$$6x^{2}+4x=4$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6x^{2}+4x}{6}=\frac{4}{6}$$

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{6}x=\frac{4}{6}$$

\frac{4}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{2}{3}x=\frac{4}{6}$$

\frac{4}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{2}{3}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2}{3} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{7}{9}$$

x^{2}+\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{7}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{\sqrt{7}}{3} x+\frac{1}{3}=-\frac{\sqrt{7}}{3}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

$$x=\frac{\sqrt{7}-1}{3} x=\frac{-\sqrt{7}-1}{3}$$