6x^2+8x-12=0 eching

$$6 x ^ { 2 } + 8 x - 12 = 0$$

x uchun yechish

$$x=\frac{\sqrt{22}-2}{3}\approx 0,896805253$$

$$x=\frac{-\sqrt{22}-2}{3}\approx -2,230138587$$

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.

$$6x^{2}+8x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, 8 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 6\left(-12\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-24\left(-12\right)}}{2\times 6}$$

-24 ni -12 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+288}}{2\times 6}$$

64 ni 288 ga qo'shish.

$$x=\frac{-8±\sqrt{352}}{2\times 6}$$

352 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±4\sqrt{22}}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{-8±4\sqrt{22}}{12}$$

x=\frac{-8±4\sqrt{22}}{12} tenglamasini yeching, bunda ± musbat. -8 ni 4\sqrt{22} ga qo'shish.

$$x=\frac{4\sqrt{22}-8}{12}$$

-8+4\sqrt{22} ni 12 ga bo'lish.

$$x=\frac{\sqrt{22}-2}{3}$$

x=\frac{-8±4\sqrt{22}}{12} tenglamasini yeching, bunda ± manfiy. -8 dan 4\sqrt{22} ni ayirish.

$$x=\frac{-4\sqrt{22}-8}{12}$$

-8-4\sqrt{22} ni 12 ga bo'lish.

$$x=\frac{-\sqrt{22}-2}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{22}-2}{3} x=\frac{-\sqrt{22}-2}{3}$$

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.

$$6x^{2}+8x-12=0$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$6x^{2}+8x-12-\left(-12\right)=-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$6x^{2}+8x=-\left(-12\right)$$

0 dan -12 ni ayirish.

$$6x^{2}+8x=12$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6x^{2}+8x}{6}=\frac{12}{6}$$

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

$$x^{2}+\frac{8}{6}x=\frac{12}{6}$$

\frac{8}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{4}{3}x=\frac{12}{6}$$

12 ni 6 ga bo'lish.

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

\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{3} olish uchun. Keyin, \frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{3} kvadratini chiqarish.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=2+\frac{4}{9}$$

2 ni \frac{4}{9} ga qo'shish.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{22}{9}$$

x^{2}+\frac{4}{3}x+\frac{4}{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.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{22}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{22}{9}}$$

Qisqartirish.

$$x+\frac{2}{3}=\frac{\sqrt{22}}{3} x+\frac{2}{3}=-\frac{\sqrt{22}}{3}$$

Tenglamaning ikkala tarafidan \frac{2}{3} ni ayirish.

$$x=\frac{\sqrt{22}-2}{3} x=\frac{-\sqrt{22}-2}{3}$$