k^2-11k-102 eching

$$k ^ { 2 } - 11 k - 102$$

Omil

$$\left(k-17\right)\left(k+6\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda k^{2}+ak+bk-102 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=1\left(-102\right)=-102$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -102-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-102 2,-51 3,-34 6,-17$$

Har bir juftlik yigʻindisini hisoblang.

$$1-102=-101 2-51=-49 3-34=-31 6-17=-11$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-17 b=6$$

k^{2}-11k-102 ni \left(k^{2}-17k\right)+\left(6k-102\right) sifatida qaytadan yozish.

$$\left(k^{2}-17k\right)+\left(6k-102\right)$$

Birinchi guruhda k ni va ikkinchi guruhda 6 ni faktordan chiqaring.

$$k\left(k-17\right)+6\left(k-17\right)$$

Distributiv funktsiyasidan foydalangan holda k-17 umumiy terminini chiqaring.

$$\left(k-17\right)\left(k+6\right)$$

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.

$$k^{2}-11k-102=0$$

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.

$$k=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-102\right)}}{2}$$

-11 kvadratini chiqarish.

$$k=\frac{-\left(-11\right)±\sqrt{121-4\left(-102\right)}}{2}$$

-4 ni -102 marotabaga ko'paytirish.

$$k=\frac{-\left(-11\right)±\sqrt{121+408}}{2}$$

121 ni 408 ga qo'shish.

$$k=\frac{-\left(-11\right)±\sqrt{529}}{2}$$

529 ning kvadrat ildizini chiqarish.

$$k=\frac{-\left(-11\right)±23}{2}$$

-11 ning teskarisi 11 ga teng.

$$k=\frac{11±23}{2}$$

k=\frac{11±23}{2} tenglamasini yeching, bunda ± musbat. 11 ni 23 ga qo'shish.

$$k=\frac{34}{2}$$

34 ni 2 ga bo'lish.

$$k=17$$

k=\frac{11±23}{2} tenglamasini yeching, bunda ± manfiy. 11 dan 23 ni ayirish.

$$k=-\frac{12}{2}$$

-12 ni 2 ga bo'lish.

$$k=-6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 17 ga va x_{2} uchun -6 ga bo‘ling.

$$k^{2}-11k-102=\left(k-17\right)\left(k-\left(-6\right)\right)$$

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

$$k^{2}-11k-102=\left(k-17\right)\left(k+6\right)$$

Baholash

$$\left(k-17\right)\left(k+6\right)$$