Matematika

8-sinf · Algebra · 25 daqiqa

To'liqsiz kvadrat tenglama

Kvadrat tenglamalar: mavzular

Bu nima va nega kerak

7-sinfda chiziqli tenglamalarni (3x+6=03x + 6 = 0) o'rgandik. Ammo maydon, tezlanish yoki erkin tushish masalalarida noma'lum ikkinchi darajada qatnashadi:

Tomoni xx bo'lgan kvadrat maydonning yuzi 64 m264\text{ m}^2 ga teng. Maydon tomonini toping.

Matematik tilda: x2=64x^2 = 64 yoki x2−64=0x^2 - 64 = 0. Bu yerda xx ikkinchi darajada turibdi. Bunday tenglamalar kvadrat tenglamalar deyiladi.

Kvadrat tenglamalar orasida umumiy formulalarsiz, ko'paytuvchilarga ajratish yoki arifmetik ildiz chiqarish bilan yechiladigan turi bor — bular to'liqsiz kvadrat tenglamalardir.

Asosiy g'oya

Ta'rif. ax2+bx+c=0ax^2 + bx + c = 0 ko'rinishidagi tenglama kvadrat tenglama deyiladi. Bu yerda:

  • xx — noma'lum;
  • a,b,ca, b, c — berilgan sonlar (koeffitsiyentlar);
  • a≠0a \neq 0 bo'lishi shart (a=0a = 0 bo'lsa, bx+c=0bx + c = 0 chiziqli tenglama qoladi);
  • aa — bosh koeffitsiyent, bb — ikkinchi koeffitsiyent, cc — ozod had.

Agar b=0b = 0 yoki c=0c = 0 bo'lsa, tenglama to'liqsiz kvadrat tenglama deyiladi.

To'liqsiz kvadrat tenglamalarning 3 ta asosiy turi bor:

TuriShartKo'rinishiYechish usuliIldizlari
1-turb=0,  c=0b = 0,\; c = 0ax2=0ax^2 = 0aa ga bo'lishx=0x = 0 (yagona)
2-turb=0,  c≠0b = 0,\; c \neq 0ax2+c=0ax^2 + c = 0x2=−cax^2 = -\dfrac{c}{a}−ca>0-\dfrac{c}{a} > 0 da: x=±−cax = \pm\sqrt{-\dfrac{c}{a}}
−ca<0-\dfrac{c}{a} < 0 da: ildiz yo'q
3-turb≠0,  c=0b \neq 0,\; c = 0ax2+bx=0ax^2 + bx = 0x(ax+b)=0x(ax + b) = 0Har doim 2 ta: x1=0,  x2=−bax_1 = 0,\; x_2 = -\dfrac{b}{a}

1-tur: ax2=0ax^2 = 0

a≠0a \neq 0 bo'lgani uchun x2=0x^2 = 0, bundan x=0x = 0.

2-tur: ax2+c=0ax^2 + c = 0 (b=0b = 0)

Ozod hadni o'ngga o'tkazamiz: ax2=−c  ⟹  x2=−caax^2 = -c \implies x^2 = -\dfrac{c}{a}.

  • Agar −ca>0-\dfrac{c}{a} > 0 bo'lsa, ikkita ildiz: x=±−cax = \pm\sqrt{-\dfrac{c}{a}}.
  • Agar −ca<0-\dfrac{c}{a} < 0 bo'lsa, sonning kvadrati manfiy bo'lolmaydi, shuning uchun haqiqiy ildiz yo'q.

3-tur: ax2+bx=0ax^2 + bx = 0 (c=0c = 0)

Umumiy ko'paytuvchi xx ni qavsdan chiqaramiz: x(ax+b)=0  ⟹  x1=0yokiax+b=0  ⟹  x2=−ba.x(ax + b) = 0 \implies x_1 = 0 \quad \text{yoki} \quad ax + b = 0 \implies x_2 = -\frac{b}{a}. Bunday tenglamada har doim ildizlardan biri nolga teng bo'ladi.

Yechilgan misollar

1-misol. b=0,c=0b = 0, c = 0 holati

5x2=0.5x^2 = 0.

  • Tenglamaning ikkala tomonini 55 ga bo'lamiz: x2=0.x^2 = 0.
  • Kvadrati 0 ga teng son faqat 0: x=0.x = 0.

Javob: x=0x = 0.

2-misol. b=0,c≠0b = 0, c \neq 0 holati (ildizlar mavjud)

3x2−48=0.3x^2 - 48 = 0.

  • Ozod had −48-48 ni o'ng tomonga ishorasini o'zgartirib o'tkazamiz: 3x2=48.3x^2 = 48.
  • Ikkala tomonni 33 ga bo'lamiz: x2=16.x^2 = 16.
  • Qaysi sonlarning kvadrati 1616 ga teng? Bu 44 va −4-4: x=±16  ⟹  x1=4,x2=−4.x = \pm\sqrt{16} \implies x_1 = 4, \quad x_2 = -4.

Tekshirish:

  • x=4x = 4 bo'lganda: 3⋅42−48=3⋅16−48=48−48=03\cdot 4^2 - 48 = 3\cdot 16 - 48 = 48 - 48 = 0 ✓.
  • x=−4x = -4 bo'lganda: 3⋅(−4)2−48=3⋅16−48=03\cdot (-4)^2 - 48 = 3\cdot 16 - 48 = 0 ✓.

Javob: x1=4,  x2=−4x_1 = 4,\; x_2 = -4.

3-misol. b=0,c≠0b = 0, c \neq 0 holati (haqiqiy ildiz yo'q)

2x2+50=0.2x^2 + 50 = 0.

  • Ozod hadni o'ng tomonga o'tkazamiz: 2x2=−50.2x^2 = -50.
  • Ikkala tomonni 22 ga bo'lamiz: x2=−25.x^2 = -25.
  • Hech qanday haqiqiy sonning kvadrati manfiy songa (−25-25 ga) teng bo'la olmaydi.

Javob: Haqiqiy ildizga ega emas (ildizi yo'q).

4-misol. c=0,b≠0c = 0, b \neq 0 holati

4x2−20x=0.4x^2 - 20x = 0.

  • Chap tomondan 4x4x ni umumiy ko'paytuvchi sifatida qavsdan chiqaramiz: 4x(x−5)=0.4x(x - 5) = 0.
  • Ko'paytuvchilarni nolga tenglashtiramiz: 4x=0  ⟹  x1=0,4x = 0 \implies x_1 = 0, x−5=0  ⟹  x2=5.x - 5 = 0 \implies x_2 = 5.

Tekshirish:

  • x=0x = 0: 4⋅02−20⋅0=04\cdot 0^2 - 20\cdot 0 = 0 ✓.
  • x=5x = 5: 4⋅52−20⋅5=4⋅25−100=04\cdot 5^2 - 20\cdot 5 = 4\cdot 25 - 100 = 0 ✓.

Javob: x1=0,  x2=5x_1 = 0,\; x_2 = 5.

5-misol. Standart ko'rinishga keltirib yechish

(x−3)(x+3)+2x=2x+7.(x - 3)(x + 3) + 2x = 2x + 7.

  • Chap tomondagi (x−3)(x+3)(x - 3)(x + 3) ifodaga kvadratlar ayirmasi formulasini qo'llaymiz: x2−9+2x=2x+7.x^2 - 9 + 2x = 2x + 7.
  • Barcha hadlarni chap tomonga o'tkazib, o'xshash hadlarni ixchamlaymiz: x2−9+2x−2x−7=0,x^2 - 9 + 2x - 2x - 7 = 0, x2−16=0.x^2 - 16 = 0.
  • Bu 2-tur to'liqsiz kvadrat tenglama: x2=16  ⟹  x=±4.x^2 = 16 \implies x = \pm 4.

Javob: x1=−4,  x2=4x_1 = -4,\; x_2 = 4.

Mashq

Quyidagi tenglamalarni yeching:

  1. x2−49=0x^2 - 49 = 0
  2. 7x2=07x^2 = 0
  3. 2x2−18x=02x^2 - 18x = 0
  4. 3x2+27=03x^2 + 27 = 0
  5. 5x2−20=05x^2 - 20 = 0
  6. 6x2+12x=06x^2 + 12x = 0
  7. (x−4)(x+4)=9(x - 4)(x + 4) = 9
  8. 4x2−7x=x2+2x4x^2 - 7x = x^2 + 2x
  9. x2−14=6\dfrac{x^2 - 1}{4} = 6
  10. (2x−1)2=−4x+17(2x - 1)^2 = -4x + 17
Javoblar va yechim yo'llari
  1. x2=49  ⟹  x=±7x^2 = 49 \implies x = \pm 7. Javob: x1=−7,  x2=7x_1 = -7,\; x_2 = 7.
  2. x2=0  ⟹  x=0x^2 = 0 \implies x = 0. Javob: x=0x = 0.
  3. 2x(x−9)=0  ⟹  x1=0,  x2=92x(x - 9) = 0 \implies x_1 = 0,\; x_2 = 9. Javob: x1=0,  x2=9x_1 = 0,\; x_2 = 9.
  4. 3x2=−27  ⟹  x2=−93x^2 = -27 \implies x^2 = -9. Haqiqiy sonning kvadrati manfiy bo'la olmaydi. Javob: Ildizga ega emas.
  5. 5x2=20  ⟹  x2=4  ⟹  x=±25x^2 = 20 \implies x^2 = 4 \implies x = \pm 2. Javob: x1=−2,  x2=2x_1 = -2,\; x_2 = 2.
  6. 6x(x+2)=0  ⟹  x1=0,  x2=−26x(x + 2) = 0 \implies x_1 = 0,\; x_2 = -2. Javob: x1=0,  x2=−2x_1 = 0,\; x_2 = -2.
  7. x2−16=9  ⟹  x2=25  ⟹  x=±5x^2 - 16 = 9 \implies x^2 = 25 \implies x = \pm 5. Javob: x1=−5,  x2=5x_1 = -5,\; x_2 = 5.
  8. Hadlarni chapga o'tkazamiz: 4x2−x2−7x−2x=0  ⟹  3x2−9x=0  ⟹  3x(x−3)=04x^2 - x^2 - 7x - 2x = 0 \implies 3x^2 - 9x = 0 \implies 3x(x - 3) = 0. Javob: x1=0,  x2=3x_1 = 0,\; x_2 = 3.
  9. Ikkala tomonni 44 ga ko'paytiramiz: x2−1=24  ⟹  x2=25  ⟹  x=±5x^2 - 1 = 24 \implies x^2 = 25 \implies x = \pm 5. Javob: x1=−5,  x2=5x_1 = -5,\; x_2 = 5.
  10. Qavsni ochamiz: 4x2−4x+1=−4x+174x^2 - 4x + 1 = -4x + 17. Hadlarni o'tkazsak: 4x2=16  ⟹  x2=4  ⟹  x=±24x^2 = 16 \implies x^2 = 4 \implies x = \pm 2. Javob: x1=−2,  x2=2x_1 = -2,\; x_2 = 2.

Keng tarqalgan xatolar

1. Noma'lum xx ga bo'lib yuborish va nol ildizni yo'qotish. ✗ 3x2−6x=0  ⟹  3x=6  ⟹  x=23x^2 - 6x = 0 \implies 3x = 6 \implies x = 2 (bitta ildiz yo'qotildi!) ✓ Har doim umumiy ko'paytuvchini qavsdan chiqaring: 3x(x−2)=0  ⟹  x1=0,x2=23x(x - 2) = 0 \implies x_1 = 0, x_2 = 2.

2. Kvadrat ildiz chiqarayotganda manfiy ildizni unutish. ✗ x2=25  ⟹  x=5x^2 = 25 \implies x = 5 ✓ Kvadrati 2525 ga teng son ikkita: x=±5x = \pm 5 (ya'ni x1=5,x2=−5x_1 = 5, x_2 = -5).

3. Manfiy songa teng bo'lganda ildiz chiqarishga urinish. ✗ x2=−16  ⟹  x=±4x^2 = -16 \implies x = \pm 4 ✓ Haqiqiy sonlar sohasida manfiy sondan kvadrat ildiz chiqmaydi: haqiqiy ildiz yo'q.

4. Kvadrat tenglamada bosh koeffitsiyentni nolga teng deb olish. ✗ 0⋅x2+4x−8=00\cdot x^2 + 4x - 8 = 0 — kvadrat tenglama deb hisoblash. ✓ Kvadrat tenglamada a≠0a \neq 0 bo'lishi shart. Agar a=0a = 0 bo'lsa, u 4x−8=04x - 8 = 0 chiziqli tenglamadir.

DTM uslubidagi test

Ushbu mavzu bo'yicha tayyorlangan 5 ta test savoli platformada interaktiv tarzda ishlaydi: o'z javobingizni tanlang va batafsil yechimni tekshiring.

Keyingi qadam

To'liqsiz kvadrat tenglamalarni yechishni o'rgandik. Ammo barcha koeffitsiyentlari noldan farqli bo'lgan to'liq kvadrat tenglamalarni (ax2+bx+c=0ax^2 + bx + c = 0) qanday yechamiz? Keyingi darsda bu muammoni hal qiluvchi universal qurol — Diskriminant va kvadrat tenglama ildizlari formulasi bilan tanishamiz.

Test

0 / 5 javob berildi

  1. 1.2x2−32=02x^2 - 32 = 0 tenglamaning ildizlarini toping.

  2. 2.3x2+12x=03x^2 + 12x = 0 tenglamaning ildizlari yig'indisini toping.

  3. 3.5x2+20=05x^2 + 20 = 0 tenglama haqida qaysi fikr to'g'ri?

  4. 4.(x−3)(x+3)=7(x - 3)(x + 3) = 7 tenglamaning musbat ildizini toping.

  5. 5.4x2−5x=04x^2 - 5x = 0 tenglamaning noldan farqli ildizini toping.