26=a^2-10a+25+4a^2-12a+9 eching

$$26 = a ^ { 2 } - 10 a + 25 + 4 a ^ { 2 } - 12 a + 9$$

a uchun yechish

$$a=\frac{2}{5}=0,4$$

$$a=4$$

Guruhlash orqali faktorlash qadamlari

5a^{2} ni olish uchun a^{2} va 4a^{2} ni birlashtirish.

$$26=5a^{2}-10a+25-12a+9$$

-22a ni olish uchun -10a va -12a ni birlashtirish.

$$26=5a^{2}-22a+25+9$$

34 olish uchun 25 va 9'ni qo'shing.

$$26=5a^{2}-22a+34$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5a^{2}-22a+34=26$$

Ikkala tarafdan 26 ni ayirish.

$$5a^{2}-22a+34-26=0$$

8 olish uchun 34 dan 26 ni ayirish.

$$5a^{2}-22a+8=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5a^{2}+aa+ba+8 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-22 ab=5\times 8=40$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 40-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

Yechim – -22 yigʻindisini beruvchi juftlik.

$$a=-20 b=-2$$

5a^{2}-22a+8 ni \left(5a^{2}-20a\right)+\left(-2a+8\right) sifatida qaytadan yozish.

$$\left(5a^{2}-20a\right)+\left(-2a+8\right)$$

Birinchi guruhda 5a ni va ikkinchi guruhda -2 ni faktordan chiqaring.

$$5a\left(a-4\right)-2\left(a-4\right)$$

Distributiv funktsiyasidan foydalangan holda a-4 umumiy terminini chiqaring.

$$\left(a-4\right)\left(5a-2\right)$$

Tenglamani yechish uchun a-4=0 va 5a-2=0 ni yeching.

$$a=4 a=\frac{2}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

5a^{2} ni olish uchun a^{2} va 4a^{2} ni birlashtirish.

$$26=5a^{2}-10a+25-12a+9$$

-22a ni olish uchun -10a va -12a ni birlashtirish.

$$26=5a^{2}-22a+25+9$$

34 olish uchun 25 va 9'ni qo'shing.

$$26=5a^{2}-22a+34$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5a^{2}-22a+34=26$$

Ikkala tarafdan 26 ni ayirish.

$$5a^{2}-22a+34-26=0$$

8 olish uchun 34 dan 26 ni ayirish.

$$5a^{2}-22a+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -22 ni b va 8 ni c bilan almashtiring.

$$a=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 5\times 8}}{2\times 5}$$

-22 kvadratini chiqarish.

$$a=\frac{-\left(-22\right)±\sqrt{484-4\times 5\times 8}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$a=\frac{-\left(-22\right)±\sqrt{484-20\times 8}}{2\times 5}$$

-20 ni 8 marotabaga ko'paytirish.

$$a=\frac{-\left(-22\right)±\sqrt{484-160}}{2\times 5}$$

484 ni -160 ga qo'shish.

$$a=\frac{-\left(-22\right)±\sqrt{324}}{2\times 5}$$

324 ning kvadrat ildizini chiqarish.

$$a=\frac{-\left(-22\right)±18}{2\times 5}$$

-22 ning teskarisi 22 ga teng.

$$a=\frac{22±18}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$a=\frac{22±18}{10}$$

a=\frac{22±18}{10} tenglamasini yeching, bunda ± musbat. 22 ni 18 ga qo'shish.

$$a=\frac{40}{10}$$

40 ni 10 ga bo'lish.

$$a=4$$

a=\frac{22±18}{10} tenglamasini yeching, bunda ± manfiy. 22 dan 18 ni ayirish.

$$a=\frac{4}{10}$$

\frac{4}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$a=\frac{2}{5}$$

Tenglama yechildi.

$$a=4 a=\frac{2}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

5a^{2} ni olish uchun a^{2} va 4a^{2} ni birlashtirish.

$$26=5a^{2}-10a+25-12a+9$$

-22a ni olish uchun -10a va -12a ni birlashtirish.

$$26=5a^{2}-22a+25+9$$

34 olish uchun 25 va 9'ni qo'shing.

$$26=5a^{2}-22a+34$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5a^{2}-22a+34=26$$

Ikkala tarafdan 34 ni ayirish.

$$5a^{2}-22a=26-34$$

-8 olish uchun 26 dan 34 ni ayirish.

$$5a^{2}-22a=-8$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5a^{2}-22a}{5}=-\frac{8}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$a^{2}-\frac{22}{5}a=-\frac{8}{5}$$

-\frac{22}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{5} olish uchun. Keyin, -\frac{11}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$a^{2}-\frac{22}{5}a+\left(-\frac{11}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{11}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{5} kvadratini chiqarish.

$$a^{2}-\frac{22}{5}a+\frac{121}{25}=-\frac{8}{5}+\frac{121}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{8}{5} ni \frac{121}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$a^{2}-\frac{22}{5}a+\frac{121}{25}=\frac{81}{25}$$

a^{2}-\frac{22}{5}a+\frac{121}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(a-\frac{11}{5}\right)^{2}=\frac{81}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(a-\frac{11}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Qisqartirish.

$$a-\frac{11}{5}=\frac{9}{5} a-\frac{11}{5}=-\frac{9}{5}$$

\frac{11}{5} ni tenglamaning ikkala tarafiga qo'shish.

$$a=4 a=\frac{2}{5}$$