12x^2-160x+400=0 eching

$$12 { x }^{ 2 } -160x+400 = 0$$

x uchun yechish

$$x = \frac{10}{3} = 3\frac{1}{3} \approx 3,333333333$$

$$x=10$$

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.

$$12x^{2}-160x+400=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 12 ni a, -160 ni b va 400 ni c bilan almashtiring.

$$x=\frac{-\left(-160\right)±\sqrt{\left(-160\right)^{2}-4\times 12\times 400}}{2\times 12}$$

-160 kvadratini chiqarish.

$$x=\frac{-\left(-160\right)±\sqrt{25600-4\times 12\times 400}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$x=\frac{-\left(-160\right)±\sqrt{25600-48\times 400}}{2\times 12}$$

-48 ni 400 marotabaga ko'paytirish.

$$x=\frac{-\left(-160\right)±\sqrt{25600-19200}}{2\times 12}$$

25600 ni -19200 ga qo'shish.

$$x=\frac{-\left(-160\right)±\sqrt{6400}}{2\times 12}$$

6400 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-160\right)±80}{2\times 12}$$

-160 ning teskarisi 160 ga teng.

$$x=\frac{160±80}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$x=\frac{160±80}{24}$$

x=\frac{160±80}{24} tenglamasini yeching, bunda ± musbat. 160 ni 80 ga qo'shish.

$$x=\frac{240}{24}$$

240 ni 24 ga bo'lish.

$$x=10$$

x=\frac{160±80}{24} tenglamasini yeching, bunda ± manfiy. 160 dan 80 ni ayirish.

$$x=\frac{80}{24}$$

\frac{80}{24} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{10}{3}$$

Tenglama yechildi.

$$x=10 x=\frac{10}{3}$$

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.

$$12x^{2}-160x+400=0$$

Tenglamaning ikkala tarafidan 400 ni ayirish.

$$12x^{2}-160x+400-400=-400$$

O‘zidan 400 ayirilsa 0 qoladi.

$$12x^{2}-160x=-400$$

Ikki tarafini 12 ga bo‘ling.

$$\frac{12x^{2}-160x}{12}=-\frac{400}{12}$$

12 ga bo'lish 12 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{160}{12}\right)x=-\frac{400}{12}$$

\frac{-160}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{40}{3}x=-\frac{400}{12}$$

\frac{-400}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{40}{3}x=-\frac{100}{3}$$

-\frac{40}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{20}{3} olish uchun. Keyin, -\frac{20}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{40}{3}x+\left(-\frac{20}{3}\right)^{2}=-\frac{100}{3}+\left(-\frac{20}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{20}{3} kvadratini chiqarish.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=-\frac{100}{3}+\frac{400}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{100}{3} ni \frac{400}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=\frac{100}{9}$$

x^{2}-\frac{40}{3}x+\frac{400}{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.

$$\left(x-\frac{20}{3}\right)^{2}=\frac{100}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{20}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Qisqartirish.

$$x-\frac{20}{3}=\frac{10}{3} x-\frac{20}{3}=-\frac{10}{3}$$

\frac{20}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=10 x=\frac{10}{3}$$