32x^2+250x-1925=0 eching

$$32 { x }^{ 2 } +250x-1925=0$$

x uchun yechish

$$x = \frac{5 \sqrt{3089} - 125}{32} \approx 4,77793327$$

$$x=\frac{-5\sqrt{3089}-125}{32}\approx -12,59043327$$

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.

$$32x^{2}+250x-1925=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 32 ni a, 250 ni b va -1925 ni c bilan almashtiring.

$$x=\frac{-250±\sqrt{250^{2}-4\times 32\left(-1925\right)}}{2\times 32}$$

250 kvadratini chiqarish.

$$x=\frac{-250±\sqrt{62500-4\times 32\left(-1925\right)}}{2\times 32}$$

-4 ni 32 marotabaga ko'paytirish.

$$x=\frac{-250±\sqrt{62500-128\left(-1925\right)}}{2\times 32}$$

-128 ni -1925 marotabaga ko'paytirish.

$$x=\frac{-250±\sqrt{62500+246400}}{2\times 32}$$

62500 ni 246400 ga qo'shish.

$$x=\frac{-250±\sqrt{308900}}{2\times 32}$$

308900 ning kvadrat ildizini chiqarish.

$$x=\frac{-250±10\sqrt{3089}}{2\times 32}$$

2 ni 32 marotabaga ko'paytirish.

$$x=\frac{-250±10\sqrt{3089}}{64}$$

x=\frac{-250±10\sqrt{3089}}{64} tenglamasini yeching, bunda ± musbat. -250 ni 10\sqrt{3089} ga qo'shish.

$$x=\frac{10\sqrt{3089}-250}{64}$$

-250+10\sqrt{3089} ni 64 ga bo'lish.

$$x=\frac{5\sqrt{3089}-125}{32}$$

x=\frac{-250±10\sqrt{3089}}{64} tenglamasini yeching, bunda ± manfiy. -250 dan 10\sqrt{3089} ni ayirish.

$$x=\frac{-10\sqrt{3089}-250}{64}$$

-250-10\sqrt{3089} ni 64 ga bo'lish.

$$x=\frac{-5\sqrt{3089}-125}{32}$$

Tenglama yechildi.

$$x=\frac{5\sqrt{3089}-125}{32} x=\frac{-5\sqrt{3089}-125}{32}$$

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.

$$32x^{2}+250x-1925=0$$

1925 ni tenglamaning ikkala tarafiga qo'shish.

$$32x^{2}+250x-1925-\left(-1925\right)=-\left(-1925\right)$$

O‘zidan -1925 ayirilsa 0 qoladi.

$$32x^{2}+250x=-\left(-1925\right)$$

0 dan -1925 ni ayirish.

$$32x^{2}+250x=1925$$

Ikki tarafini 32 ga bo‘ling.

$$\frac{32x^{2}+250x}{32}=\frac{1925}{32}$$

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

$$x^{2}+\frac{250}{32}x=\frac{1925}{32}$$

\frac{250}{32} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{125}{16}x=\frac{1925}{32}$$

\frac{125}{16} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{125}{32} olish uchun. Keyin, \frac{125}{32} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{125}{16}x+\left(\frac{125}{32}\right)^{2}=\frac{1925}{32}+\left(\frac{125}{32}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{125}{32} kvadratini chiqarish.

$$x^{2}+\frac{125}{16}x+\frac{15625}{1024}=\frac{1925}{32}+\frac{15625}{1024}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1925}{32} ni \frac{15625}{1024} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{125}{16}x+\frac{15625}{1024}=\frac{77225}{1024}$$

x^{2}+\frac{125}{16}x+\frac{15625}{1024} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{125}{32}\right)^{2}=\frac{77225}{1024}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{125}{32}\right)^{2}}=\sqrt{\frac{77225}{1024}}$$

Qisqartirish.

$$x+\frac{125}{32}=\frac{5\sqrt{3089}}{32} x+\frac{125}{32}=-\frac{5\sqrt{3089}}{32}$$

Tenglamaning ikkala tarafidan \frac{125}{32} ni ayirish.

$$x=\frac{5\sqrt{3089}-125}{32} x=\frac{-5\sqrt{3089}-125}{32}$$