2x^2-11x-40=0 eching

$$2 { x }^{ 2 } -11x-40=0$$

x uchun yechish

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

$$x=8$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2x^{2}+ax+bx-40 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=2\left(-40\right)=-80$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -80-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-16 b=5$$

2x^{2}-11x-40 ni \left(2x^{2}-16x\right)+\left(5x-40\right) sifatida qaytadan yozish.

$$\left(2x^{2}-16x\right)+\left(5x-40\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$2x\left(x-8\right)+5\left(x-8\right)$$

Distributiv funktsiyasidan foydalangan holda x-8 umumiy terminini chiqaring.

$$\left(x-8\right)\left(2x+5\right)$$

Tenglamani yechish uchun x-8=0 va 2x+5=0 ni yeching.

$$x=8 x=-\frac{5}{2}$$

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.

$$2x^{2}-11x-40=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -11 ni b va -40 ni c bilan almashtiring.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

-11 kvadratini chiqarish.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-40\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-40\right)}}{2\times 2}$$

-8 ni -40 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121+320}}{2\times 2}$$

121 ni 320 ga qo'shish.

$$x=\frac{-\left(-11\right)±\sqrt{441}}{2\times 2}$$

441 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-11\right)±21}{2\times 2}$$

-11 ning teskarisi 11 ga teng.

$$x=\frac{11±21}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{11±21}{4}$$

x=\frac{11±21}{4} tenglamasini yeching, bunda ± musbat. 11 ni 21 ga qo'shish.

$$x=\frac{32}{4}$$

32 ni 4 ga bo'lish.

$$x=8$$

x=\frac{11±21}{4} tenglamasini yeching, bunda ± manfiy. 11 dan 21 ni ayirish.

$$x=-\frac{10}{4}$$

\frac{-10}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=8 x=-\frac{5}{2}$$

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.

$$2x^{2}-11x-40=0$$

40 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}-11x-40-\left(-40\right)=-\left(-40\right)$$

O‘zidan -40 ayirilsa 0 qoladi.

$$2x^{2}-11x=-\left(-40\right)$$

0 dan -40 ni ayirish.

$$2x^{2}-11x=40$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-11x}{2}=\frac{40}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{11}{2}x=\frac{40}{2}$$

40 ni 2 ga bo'lish.

$$x^{2}-\frac{11}{2}x=20$$

-\frac{11}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{4} olish uchun. Keyin, -\frac{11}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=20+\left(-\frac{11}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{4} kvadratini chiqarish.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=20+\frac{121}{16}$$

20 ni \frac{121}{16} ga qo'shish.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{441}{16}$$

x^{2}-\frac{11}{2}x+\frac{121}{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.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{441}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{441}{16}}$$

Qisqartirish.

$$x-\frac{11}{4}=\frac{21}{4} x-\frac{11}{4}=-\frac{21}{4}$$

\frac{11}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=8 x=-\frac{5}{2}$$