25x^2-10x-x^2-25=0 eching

$$25 x ^ { 2 } - 10 x - x ^ { 2 } - 25 = 0$$

x uchun yechish

$$x=-\frac{5}{6}\approx -0,833333333$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1,25$$

Guruhlash orqali faktorlash qadamlari

24x^{2} ni olish uchun 25x^{2} va -x^{2} ni birlashtirish.

$$24x^{2}-10x-25=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 24x^{2}+ax+bx-25 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-10 ab=24\left(-25\right)=-600$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -600-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-600 2,-300 3,-200 4,-150 5,-120 6,-100 8,-75 10,-60 12,-50 15,-40 20,-30 24,-25$$

Har bir juftlik yigʻindisini hisoblang.

$$1-600=-599 2-300=-298 3-200=-197 4-150=-146 5-120=-115 6-100=-94 8-75=-67 10-60=-50 12-50=-38 15-40=-25 20-30=-10 24-25=-1$$

Yechim – -10 yigʻindisini beruvchi juftlik.

$$a=-30 b=20$$

24x^{2}-10x-25 ni \left(24x^{2}-30x\right)+\left(20x-25\right) sifatida qaytadan yozish.

$$\left(24x^{2}-30x\right)+\left(20x-25\right)$$

Birinchi guruhda 6x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$6x\left(4x-5\right)+5\left(4x-5\right)$$

Distributiv funktsiyasidan foydalangan holda 4x-5 umumiy terminini chiqaring.

$$\left(4x-5\right)\left(6x+5\right)$$

Tenglamani yechish uchun 4x-5=0 va 6x+5=0 ni yeching.

$$x=\frac{5}{4} x=-\frac{5}{6}$$

Kvadrat tenglamalaridan foydalanish qoidalari

24x^{2} ni olish uchun 25x^{2} va -x^{2} ni birlashtirish.

$$24x^{2}-10x-25=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 24 ni a, -10 ni b va -25 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 24\left(-25\right)}}{2\times 24}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 24\left(-25\right)}}{2\times 24}$$

-4 ni 24 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-96\left(-25\right)}}{2\times 24}$$

-96 ni -25 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+2400}}{2\times 24}$$

100 ni 2400 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{2500}}{2\times 24}$$

2500 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±50}{2\times 24}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±50}{2\times 24}$$

2 ni 24 marotabaga ko'paytirish.

$$x=\frac{10±50}{48}$$

x=\frac{10±50}{48} tenglamasini yeching, bunda ± musbat. 10 ni 50 ga qo'shish.

$$x=\frac{60}{48}$$

\frac{60}{48} ulushini 12 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{10±50}{48} tenglamasini yeching, bunda ± manfiy. 10 dan 50 ni ayirish.

$$x=-\frac{40}{48}$$

\frac{-40}{48} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{5}{6}$$

Tenglama yechildi.

$$x=\frac{5}{4} x=-\frac{5}{6}$$

Kvadratni yakunlash bo'yicha qadamlar

24x^{2} ni olish uchun 25x^{2} va -x^{2} ni birlashtirish.

$$24x^{2}-10x-25=0$$

25 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$24x^{2}-10x=25$$

Ikki tarafini 24 ga bo‘ling.

$$\frac{24x^{2}-10x}{24}=\frac{25}{24}$$

24 ga bo'lish 24 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{10}{24}\right)x=\frac{25}{24}$$

\frac{-10}{24} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{12}x=\frac{25}{24}$$

-\frac{5}{12} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{24} olish uchun. Keyin, -\frac{5}{24} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{12}x+\left(-\frac{5}{24}\right)^{2}=\frac{25}{24}+\left(-\frac{5}{24}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{24} kvadratini chiqarish.

$$x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{25}{24}+\frac{25}{576}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{25}{24} ni \frac{25}{576} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{625}{576}$$

x^{2}-\frac{5}{12}x+\frac{25}{576} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{24}\right)^{2}=\frac{625}{576}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{24}\right)^{2}}=\sqrt{\frac{625}{576}}$$

Qisqartirish.

$$x-\frac{5}{24}=\frac{25}{24} x-\frac{5}{24}=-\frac{25}{24}$$

\frac{5}{24} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5}{4} x=-\frac{5}{6}$$