-2x^2-17x+30 eching

$$- 2 x ^ { 2 } - 17 x + 30$$

Omil

$$-\left(2x-3\right)\left(x+10\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -2x^{2}+ax+bx+30 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-17 ab=-2\times 30=-60$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

Yechim – -17 yigʻindisini beruvchi juftlik.

$$a=3 b=-20$$

-2x^{2}-17x+30 ni \left(-2x^{2}+3x\right)+\left(-20x+30\right) sifatida qaytadan yozish.

$$\left(-2x^{2}+3x\right)+\left(-20x+30\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -10 ni faktordan chiqaring.

$$-x\left(2x-3\right)-10\left(2x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 2x-3 umumiy terminini chiqaring.

$$\left(2x-3\right)\left(-x-10\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$-2x^{2}-17x+30=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\left(-2\right)\times 30}}{2\left(-2\right)}$$

-17 kvadratini chiqarish.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\left(-2\right)\times 30}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289+8\times 30}}{2\left(-2\right)}$$

8 ni 30 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289+240}}{2\left(-2\right)}$$

289 ni 240 ga qo'shish.

$$x=\frac{-\left(-17\right)±\sqrt{529}}{2\left(-2\right)}$$

529 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-17\right)±23}{2\left(-2\right)}$$

-17 ning teskarisi 17 ga teng.

$$x=\frac{17±23}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{17±23}{-4}$$

x=\frac{17±23}{-4} tenglamasini yeching, bunda ± musbat. 17 ni 23 ga qo'shish.

$$x=\frac{40}{-4}$$

40 ni -4 ga bo'lish.

$$x=-10$$

x=\frac{17±23}{-4} tenglamasini yeching, bunda ± manfiy. 17 dan 23 ni ayirish.

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

\frac{-6}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -10 ga va x_{2} uchun \frac{3}{2} ga bo‘ling.

$$-2x^{2}-17x+30=-2\left(x-\left(-10\right)\right)\left(x-\frac{3}{2}\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$-2x^{2}-17x+30=-2\left(x+10\right)\left(x-\frac{3}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{2} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-2x^{2}-17x+30=-2\left(x+10\right)\times \frac{-2x+3}{-2}$$

-2 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$-2x^{2}-17x+30=\left(x+10\right)\left(-2x+3\right)$$

Baholash

$$-\left(2x-3\right)\left(x+10\right)$$