-80x^2+50x-5=0 eching

$$-80 { x }^{ 2 } +50x-5=0$$

x uchun yechish

$$x=\frac{1}{8}=0,125$$

$$x=\frac{1}{2}=0,5$$

Guruhlash orqali faktorlash qadamlari

Ikki tarafini 5 ga bo‘ling.

$$-16x^{2}+10x-1=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -16x^{2}+ax+bx-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=10 ab=-16\left(-1\right)=16$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 16-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,16 2,8 4,4$$

Har bir juftlik yigʻindisini hisoblang.

$$1+16=17 2+8=10 4+4=8$$

Yechim – 10 yigʻindisini beruvchi juftlik.

$$a=8 b=2$$

-16x^{2}+10x-1 ni \left(-16x^{2}+8x\right)+\left(2x-1\right) sifatida qaytadan yozish.

$$\left(-16x^{2}+8x\right)+\left(2x-1\right)$$

-16x^{2}+8x ichida -8x ni ajrating.

$$-8x\left(2x-1\right)+2x-1$$

Distributiv funktsiyasidan foydalangan holda 2x-1 umumiy terminini chiqaring.

$$\left(2x-1\right)\left(-8x+1\right)$$

Tenglamani yechish uchun 2x-1=0 va -8x+1=0 ni yeching.

$$x=\frac{1}{2} x=\frac{1}{8}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$-80x^{2}+50x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -80 ni a, 50 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-80\right)\left(-5\right)}}{2\left(-80\right)}$$

50 kvadratini chiqarish.

$$x=\frac{-50±\sqrt{2500-4\left(-80\right)\left(-5\right)}}{2\left(-80\right)}$$

-4 ni -80 marotabaga ko'paytirish.

$$x=\frac{-50±\sqrt{2500+320\left(-5\right)}}{2\left(-80\right)}$$

320 ni -5 marotabaga ko'paytirish.

$$x=\frac{-50±\sqrt{2500-1600}}{2\left(-80\right)}$$

2500 ni -1600 ga qo'shish.

$$x=\frac{-50±\sqrt{900}}{2\left(-80\right)}$$

900 ning kvadrat ildizini chiqarish.

$$x=\frac{-50±30}{2\left(-80\right)}$$

2 ni -80 marotabaga ko'paytirish.

$$x=\frac{-50±30}{-160}$$

x=\frac{-50±30}{-160} tenglamasini yeching, bunda ± musbat. -50 ni 30 ga qo'shish.

$$x=-\frac{20}{-160}$$

\frac{-20}{-160} ulushini 20 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{8}$$

x=\frac{-50±30}{-160} tenglamasini yeching, bunda ± manfiy. -50 dan 30 ni ayirish.

$$x=-\frac{80}{-160}$$

\frac{-80}{-160} ulushini 80 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{1}{8} x=\frac{1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-80x^{2}+50x-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$-80x^{2}+50x-5-\left(-5\right)=-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$-80x^{2}+50x=-\left(-5\right)$$

0 dan -5 ni ayirish.

$$-80x^{2}+50x=5$$

Ikki tarafini -80 ga bo‘ling.

$$\frac{-80x^{2}+50x}{-80}=\frac{5}{-80}$$

-80 ga bo'lish -80 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{50}{-80}x=\frac{5}{-80}$$

\frac{50}{-80} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{8}x=\frac{5}{-80}$$

\frac{5}{-80} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{8}x=-\frac{1}{16}$$

-\frac{5}{8} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{16} olish uchun. Keyin, -\frac{5}{16} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{1}{16}+\left(-\frac{5}{16}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{16} kvadratini chiqarish.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{1}{16}+\frac{25}{256}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{16} ni \frac{25}{256} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=\frac{9}{256}$$

x^{2}-\frac{5}{8}x+\frac{25}{256} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{16}\right)^{2}=\frac{9}{256}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{16}\right)^{2}}=\sqrt{\frac{9}{256}}$$

Qisqartirish.

$$x-\frac{5}{16}=\frac{3}{16} x-\frac{5}{16}=-\frac{3}{16}$$

\frac{5}{16} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{2} x=\frac{1}{8}$$