32x^2-80x+48=0 eching

$$32 { x }^{ 2 } -80x+48=0$$

x uchun yechish

$$x=1$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1,5$$

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.

$$32x^{2}-80x+48=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 32 ni a, -80 ni b va 48 ni c bilan almashtiring.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 32\times 48}}{2\times 32}$$

-80 kvadratini chiqarish.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 32\times 48}}{2\times 32}$$

-4 ni 32 marotabaga ko'paytirish.

$$x=\frac{-\left(-80\right)±\sqrt{6400-128\times 48}}{2\times 32}$$

-128 ni 48 marotabaga ko'paytirish.

$$x=\frac{-\left(-80\right)±\sqrt{6400-6144}}{2\times 32}$$

6400 ni -6144 ga qo'shish.

$$x=\frac{-\left(-80\right)±\sqrt{256}}{2\times 32}$$

256 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-80\right)±16}{2\times 32}$$

-80 ning teskarisi 80 ga teng.

$$x=\frac{80±16}{2\times 32}$$

2 ni 32 marotabaga ko'paytirish.

$$x=\frac{80±16}{64}$$

x=\frac{80±16}{64} tenglamasini yeching, bunda ± musbat. 80 ni 16 ga qo'shish.

$$x=\frac{96}{64}$$

\frac{96}{64} ulushini 32 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{80±16}{64} tenglamasini yeching, bunda ± manfiy. 80 dan 16 ni ayirish.

$$x=\frac{64}{64}$$

64 ni 64 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=\frac{3}{2} x=1$$

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.

$$32x^{2}-80x+48=0$$

Tenglamaning ikkala tarafidan 48 ni ayirish.

$$32x^{2}-80x+48-48=-48$$

O‘zidan 48 ayirilsa 0 qoladi.

$$32x^{2}-80x=-48$$

Ikki tarafini 32 ga bo‘ling.

$$\frac{32x^{2}-80x}{32}=-\frac{48}{32}$$

32 ga bo'lish 32 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{80}{32}\right)x=-\frac{48}{32}$$

\frac{-80}{32} ulushini 16 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{2}x=-\frac{48}{32}$$

\frac{-48}{32} ulushini 16 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{2}x=-\frac{3}{2}$$

-\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{4} olish uchun. Keyin, -\frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{4} kvadratini chiqarish.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{3}{2}+\frac{25}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{2} ni \frac{25}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{1}{16}$$

x^{2}-\frac{5}{2}x+\frac{25}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Qisqartirish.

$$x-\frac{5}{4}=\frac{1}{4} x-\frac{5}{4}=-\frac{1}{4}$$

\frac{5}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3}{2} x=1$$