5x^2=11x-2 eching

$$5 { x }^{ 2 } = 11x-2$$

x uchun yechish

$$x=\frac{1}{5}=0,2$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 11x ni ayirish.

$$5x^{2}-11x=-2$$

2 ni ikki tarafga qo’shing.

$$5x^{2}-11x+2=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5x^{2}+ax+bx+2 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=5\times 2=10$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-10 -2,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-10=-11 -2-5=-7$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-10 b=-1$$

5x^{2}-11x+2 ni \left(5x^{2}-10x\right)+\left(-x+2\right) sifatida qaytadan yozish.

$$\left(5x^{2}-10x\right)+\left(-x+2\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$5x\left(x-2\right)-\left(x-2\right)$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(5x-1\right)$$

Tenglamani yechish uchun x-2=0 va 5x-1=0 ni yeching.

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

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 11x ni ayirish.

$$5x^{2}-11x=-2$$

2 ni ikki tarafga qo’shing.

$$5x^{2}-11x+2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -11 ni b va 2 ni c bilan almashtiring.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 5\times 2}}{2\times 5}$$

-11 kvadratini chiqarish.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 5\times 2}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121-20\times 2}}{2\times 5}$$

-20 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121-40}}{2\times 5}$$

121 ni -40 ga qo'shish.

$$x=\frac{-\left(-11\right)±\sqrt{81}}{2\times 5}$$

81 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-11\right)±9}{2\times 5}$$

-11 ning teskarisi 11 ga teng.

$$x=\frac{11±9}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{11±9}{10}$$

x=\frac{11±9}{10} tenglamasini yeching, bunda ± musbat. 11 ni 9 ga qo'shish.

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

20 ni 10 ga bo'lish.

$$x=2$$

x=\frac{11±9}{10} tenglamasini yeching, bunda ± manfiy. 11 dan 9 ni ayirish.

$$x=\frac{2}{10}$$

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 11x ni ayirish.

$$5x^{2}-11x=-2$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-11x}{5}=-\frac{2}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{11}{5}x=-\frac{2}{5}$$

-\frac{11}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{10} olish uchun. Keyin, -\frac{11}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{10} kvadratini chiqarish.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{2}{5} ni \frac{121}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

x^{2}-\frac{11}{5}x+\frac{121}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Qisqartirish.

$$x-\frac{11}{10}=\frac{9}{10} x-\frac{11}{10}=-\frac{9}{10}$$

\frac{11}{10} ni tenglamaning ikkala tarafiga qo'shish.

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