3x^2-20x-12=10 eching

$$3 x ^ { 2 } - 20 x - 12 = 10$$

x uchun yechish

$$x = \frac{\sqrt{166} + 10}{3} \approx 7,628032909$$

$$x=\frac{10-\sqrt{166}}{3}\approx -0,961366242$$

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.

$$3x^{2}-20x-12=10$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$3x^{2}-20x-12-10=10-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$3x^{2}-20x-12-10=0$$

-12 dan 10 ni ayirish.

$$3x^{2}-20x-22=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -20 ni b va -22 ni c bilan almashtiring.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-22\right)}}{2\times 3}$$

-20 kvadratini chiqarish.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-22\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-22\right)}}{2\times 3}$$

-12 ni -22 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400+264}}{2\times 3}$$

400 ni 264 ga qo'shish.

$$x=\frac{-\left(-20\right)±\sqrt{664}}{2\times 3}$$

664 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-20\right)±2\sqrt{166}}{2\times 3}$$

-20 ning teskarisi 20 ga teng.

$$x=\frac{20±2\sqrt{166}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{20±2\sqrt{166}}{6}$$

x=\frac{20±2\sqrt{166}}{6} tenglamasini yeching, bunda ± musbat. 20 ni 2\sqrt{166} ga qo'shish.

$$x=\frac{2\sqrt{166}+20}{6}$$

20+2\sqrt{166} ni 6 ga bo'lish.

$$x=\frac{\sqrt{166}+10}{3}$$

x=\frac{20±2\sqrt{166}}{6} tenglamasini yeching, bunda ± manfiy. 20 dan 2\sqrt{166} ni ayirish.

$$x=\frac{20-2\sqrt{166}}{6}$$

20-2\sqrt{166} ni 6 ga bo'lish.

$$x=\frac{10-\sqrt{166}}{3}$$

Tenglama yechildi.

$$x=\frac{\sqrt{166}+10}{3} x=\frac{10-\sqrt{166}}{3}$$

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.

$$3x^{2}-20x-12=10$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$3x^{2}-20x-12-\left(-12\right)=10-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$3x^{2}-20x=10-\left(-12\right)$$

10 dan -12 ni ayirish.

$$3x^{2}-20x=22$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-20x}{3}=\frac{22}{3}$$

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

$$x^{2}-\frac{20}{3}x=\frac{22}{3}$$

-\frac{20}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{10}{3} olish uchun. Keyin, -\frac{10}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=\frac{22}{3}+\left(-\frac{10}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{10}{3} kvadratini chiqarish.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{22}{3}+\frac{100}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{22}{3} ni \frac{100}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{166}{9}$$

x^{2}-\frac{20}{3}x+\frac{100}{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.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{166}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{166}{9}}$$

Qisqartirish.

$$x-\frac{10}{3}=\frac{\sqrt{166}}{3} x-\frac{10}{3}=-\frac{\sqrt{166}}{3}$$

\frac{10}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{166}+10}{3} x=\frac{10-\sqrt{166}}{3}$$