24x^2+10x-21 eching

$$24 { x }^{ 2 } +10x-21$$

Omil

$$\left(4x-3\right)\left(6x+7\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 24x^{2}+ax+bx-21 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=10 ab=24\left(-21\right)=-504$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -504-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,504 -2,252 -3,168 -4,126 -6,84 -7,72 -8,63 -9,56 -12,42 -14,36 -18,28 -21,24$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+504=503 -2+252=250 -3+168=165 -4+126=122 -6+84=78 -7+72=65 -8+63=55 -9+56=47 -12+42=30 -14+36=22 -18+28=10 -21+24=3$$

Yechim – 10 yigʻindisini beruvchi juftlik.

$$a=-18 b=28$$

24x^{2}+10x-21 ni \left(24x^{2}-18x\right)+\left(28x-21\right) sifatida qaytadan yozish.

$$\left(24x^{2}-18x\right)+\left(28x-21\right)$$

Birinchi guruhda 6x ni va ikkinchi guruhda 7 ni faktordan chiqaring.

$$6x\left(4x-3\right)+7\left(4x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 4x-3 umumiy terminini chiqaring.

$$\left(4x-3\right)\left(6x+7\right)$$

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.

$$24x^{2}+10x-21=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 24\left(-21\right)}}{2\times 24}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 24\left(-21\right)}}{2\times 24}$$

-4 ni 24 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-96\left(-21\right)}}{2\times 24}$$

-96 ni -21 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+2016}}{2\times 24}$$

100 ni 2016 ga qo'shish.

$$x=\frac{-10±\sqrt{2116}}{2\times 24}$$

2116 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±46}{2\times 24}$$

2 ni 24 marotabaga ko'paytirish.

$$x=\frac{-10±46}{48}$$

x=\frac{-10±46}{48} tenglamasini yeching, bunda ± musbat. -10 ni 46 ga qo'shish.

$$x=\frac{36}{48}$$

\frac{36}{48} ulushini 12 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{3}{4}$$

x=\frac{-10±46}{48} tenglamasini yeching, bunda ± manfiy. -10 dan 46 ni ayirish.

$$x=-\frac{56}{48}$$

\frac{-56}{48} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{7}{6}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3}{4} ga va x_{2} uchun -\frac{7}{6} ga bo‘ling.

$$24x^{2}+10x-21=24\left(x-\frac{3}{4}\right)\left(x-\left(-\frac{7}{6}\right)\right)$$

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

$$24x^{2}+10x-21=24\left(x-\frac{3}{4}\right)\left(x+\frac{7}{6}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{4} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$24x^{2}+10x-21=24\times \frac{4x-3}{4}\left(x+\frac{7}{6}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{6} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$24x^{2}+10x-21=24\times \frac{4x-3}{4}\times \frac{6x+7}{6}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{4x-3}{4} ni \frac{6x+7}{6} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$24x^{2}+10x-21=24\times \frac{\left(4x-3\right)\left(6x+7\right)}{4\times 6}$$

4 ni 6 marotabaga ko'paytirish.

$$24x^{2}+10x-21=24\times \frac{\left(4x-3\right)\left(6x+7\right)}{24}$$

24 va 24 ichida eng katta umumiy 24 faktorini bekor qiling.

$$24x^{2}+10x-21=\left(4x-3\right)\left(6x+7\right)$$

Baholash

$$\left(4x-3\right)\left(6x+7\right)$$