w^2-11w+28 eching

$$w ^ { 2 } - 11 w + 28$$

Omil

$$\left(w-7\right)\left(w-4\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda w^{2}+aw+bw+28 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=1\times 28=28$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 28-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-28 -2,-14 -4,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-7 b=-4$$

w^{2}-11w+28 ni \left(w^{2}-7w\right)+\left(-4w+28\right) sifatida qaytadan yozish.

$$\left(w^{2}-7w\right)+\left(-4w+28\right)$$

Birinchi guruhda w ni va ikkinchi guruhda -4 ni faktordan chiqaring.

$$w\left(w-7\right)-4\left(w-7\right)$$

Distributiv funktsiyasidan foydalangan holda w-7 umumiy terminini chiqaring.

$$\left(w-7\right)\left(w-4\right)$$

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.

$$w^{2}-11w+28=0$$

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.

$$w=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}$$

-11 kvadratini chiqarish.

$$w=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}$$

-4 ni 28 marotabaga ko'paytirish.

$$w=\frac{-\left(-11\right)±\sqrt{121-112}}{2}$$

121 ni -112 ga qo'shish.

$$w=\frac{-\left(-11\right)±\sqrt{9}}{2}$$

9 ning kvadrat ildizini chiqarish.

$$w=\frac{-\left(-11\right)±3}{2}$$

-11 ning teskarisi 11 ga teng.

$$w=\frac{11±3}{2}$$

w=\frac{11±3}{2} tenglamasini yeching, bunda ± musbat. 11 ni 3 ga qo'shish.

$$w=\frac{14}{2}$$

14 ni 2 ga bo'lish.

$$w=7$$

w=\frac{11±3}{2} tenglamasini yeching, bunda ± manfiy. 11 dan 3 ni ayirish.

$$w=\frac{8}{2}$$

8 ni 2 ga bo'lish.

$$w=4$$

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

$$w^{2}-11w+28=\left(w-7\right)\left(w-4\right)$$

Baholash

$$\left(w-7\right)\left(w-4\right)$$