-2x^2-7x-3=0 eching

$$- 2 x ^ { 2 } - 7 x - 3 = 0$$

x uchun yechish

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

$$x=-3$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -2x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-7 ab=-2\left(-3\right)=6$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 6-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-6 -2,-3$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-6=-7 -2-3=-5$$

Yechim – -7 yigʻindisini beruvchi juftlik.

$$a=-1 b=-6$$

-2x^{2}-7x-3 ni \left(-2x^{2}-x\right)+\left(-6x-3\right) sifatida qaytadan yozish.

$$\left(-2x^{2}-x\right)+\left(-6x-3\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -3 ni faktordan chiqaring.

$$-x\left(2x+1\right)-3\left(2x+1\right)$$

Distributiv funktsiyasidan foydalangan holda 2x+1 umumiy terminini chiqaring.

$$\left(2x+1\right)\left(-x-3\right)$$

Tenglamani yechish uchun 2x+1=0 va -x-3=0 ni yeching.

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

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.

$$-2x^{2}-7x-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -2 ni a, -7 ni b va -3 ni c bilan almashtiring.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

-7 kvadratini chiqarish.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49+8\left(-3\right)}}{2\left(-2\right)}$$

8 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\left(-2\right)}$$

49 ni -24 ga qo'shish.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\left(-2\right)}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-7\right)±5}{2\left(-2\right)}$$

-7 ning teskarisi 7 ga teng.

$$x=\frac{7±5}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{7±5}{-4}$$

x=\frac{7±5}{-4} tenglamasini yeching, bunda ± musbat. 7 ni 5 ga qo'shish.

$$x=\frac{12}{-4}$$

12 ni -4 ga bo'lish.

$$x=-3$$

x=\frac{7±5}{-4} tenglamasini yeching, bunda ± manfiy. 7 dan 5 ni ayirish.

$$x=\frac{2}{-4}$$

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

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

Tenglama yechildi.

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

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.

$$-2x^{2}-7x-3=0$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$-2x^{2}-7x-3-\left(-3\right)=-\left(-3\right)$$

O‘zidan -3 ayirilsa 0 qoladi.

$$-2x^{2}-7x=-\left(-3\right)$$

0 dan -3 ni ayirish.

$$-2x^{2}-7x=3$$

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}-7x}{-2}=\frac{3}{-2}$$

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

$$x^{2}+\left(-\frac{7}{-2}\right)x=\frac{3}{-2}$$

-7 ni -2 ga bo'lish.

$$x^{2}+\frac{7}{2}x=\frac{3}{-2}$$

3 ni -2 ga bo'lish.

$$x^{2}+\frac{7}{2}x=-\frac{3}{2}$$

\frac{7}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{4} olish uchun. Keyin, \frac{7}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{7}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{4} kvadratini chiqarish.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{2} ni \frac{49}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

x^{2}+\frac{7}{2}x+\frac{49}{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.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Qisqartirish.

$$x+\frac{7}{4}=\frac{5}{4} x+\frac{7}{4}=-\frac{5}{4}$$

Tenglamaning ikkala tarafidan \frac{7}{4} ni ayirish.

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