Kvadrat tenglamaga keltiriladigan tenglamalar

Bu nima va nega kerak

Ko'pincha 4-darajali ($x^4 - 5x^2 + 4 = 0$) yoki noma'lum maxrajda bo'lgan kasrli tenglamalar uchraydi.

Bunday tenglamalar dastlab murakkab ko'rinsa-da, aslida yashirin kvadrat tenglamalardir. Yangi o'zgaruvchi kiritish (almashtirish) yoki umumiy maxrajga keltirish orqali ularni kvadrat tenglamaga keltirib oson yechish mumkin.

Asosiy g'oya

1. Bikvadrat tenglamalar

$ax^4 + bx^2 + c = 0$ ($a \neq 0$) ko'rinishidagi tenglama bikvadrat tenglama deyiladi.

$x^2 = t$ ($t \ge 0$) deb olsak: $at^2 + bt + c = 0.$

• $t > 0$ bo'lsa: $x = \pm\sqrt{t}$;
• $t = 0$ bo'lsa: $x = 0$;
• $t < 0$ bo'lsa: haqiqiy ildiz yo'q ($x^2 \ge 0$ bo'lgani uchun).

2. Yangi o'zgaruvchi kiritish

Bir xil ifoda takrorlansa, uni $t$ deb olamiz: $a(f(x))^2 + b(f(x)) + c = 0 \implies f(x) = t \implies at^2 + bt + c = 0.$

3. Kasr-ratsional tenglamalar

1. Aniqlanish sohasi: maxrajlar noldan farqli bo'lishi kerak ($Q(x) \neq 0$).
2. Umumiy maxrajga ko'paytirib, kvadrat tenglamaga keltiramiz va yechamiz.
3. Chet ildizlarni tekshirish: maxrajni nol qiladigan ildizlarni chiqarib tashlaymiz.

Yechilgan misollar

1-misol. Bikvadrat tenglama (4 ta ildiz)

$x^4 - 5x^2 + 4 = 0.$

• $x^2 = t$ ($t \ge 0$) desak: $t^2 - 5t + 4 = 0 \implies t_1 = 1, t_2 = 4$.
• Ikkala $t$ ham musbat:
  1. $x^2 = 1 \implies x = \pm 1$;
  2. $x^2 = 4 \implies x = \pm 2$.

Javob: $x = \pm 1,\; \pm 2$.

2-misol. Manfiy $t$ holati

$x^4 + 2x^2 - 8 = 0.$

• $x^2 = t$ ($t \ge 0$): $t^2 + 2t - 8 = 0 \implies t_1 = 2, t_2 = -4$.
• $t_1 = 2 \implies x = \pm\sqrt{2}$.
• $t_2 = -4 < 0$ da haqiqiy ildiz yo'q.

Javob: $x = \pm\sqrt{2}$.

3-misol. Ifodani almashtirish

$(x^2 - x)^2 - 8(x^2 - x) + 12 = 0.$

• $x^2 - x = t \implies t^2 - 8t + 12 = 0 \implies t_1 = 2, t_2 = 6$.
• Dastlabki o'zgaruvchiga qaytamiz:
  1. $x^2 - x = 2 \implies x^2 - x - 2 = 0 \implies x_1 = 2, x_2 = -1$;
  2. $x^2 - x = 6 \implies x^2 - x - 6 = 0 \implies x_3 = 3, x_4 = -2$.

Javob: $x_1 = -2,\; x_2 = -1,\; x_3 = 2,\; x_4 = 3$.

4-misol. Kasr-ratsional tenglama

$\frac{2x}{x - 2} - \frac{x + 1}{x + 2} = \frac{8}{x^2 - 4}.$

• Aniqlanish sohasi: $x \neq \pm 2$.
• Umumiy maxraj $(x - 2)(x + 2)$ ga ko'paytiramiz: $2x(x + 2) - (x + 1)(x - 2) = 8 \implies x^2 + 5x - 6 = 0.$
• Viyet teoremasiga ko'ra: $x_1 = 1, x_2 = -6$.
• Ikkala ildiz ham $x \neq \pm 2$ shartiga mos.

Javob: $x_1 = -6,\; x_2 = 1$.

5-misol. Chet ildizli kasr tenglama

$\frac{x^2 - 2x}{x - 2} = 3x - 4.$

• Aniqlanish sohasi: $x \neq 2$.
• Suratni ko'paytuvchiga ajratamiz: $\dfrac{x(x - 2)}{x - 2} = 3x - 4 \implies x = 3x - 4 \implies x = 2$.
• Ammo $x = 2$ da maxraj 0 bo'ladi (chet ildiz).

Javob: Haqiqiy ildizga ega emas.

Mashq

Quyidagi tenglamalarni yeching:

1. $x^4 - 10x^2 + 9 = 0$
2. $x^4 - 8x^2 - 9 = 0$
3. $x^4 - 16 = 0$
4. $(x^2 + 2x)^2 - 2(x^2 + 2x) - 3 = 0$
5. $\dfrac{x^2 - 4x}{x - 4} = 0$
6. $\dfrac{x}{x-3} - \dfrac{3}{x+3} = \dfrac{18}{x^2-9}$
7. $\dfrac{x^2}{x-1} = \dfrac{4x-3}{x-1}$
8. $x^4 + 5x^2 + 6 = 0$
9. $\left(\dfrac{x-1}{x}\right)^2 - 3\left(\dfrac{x-1}{x}\right) + 2 = 0$
10. $\dfrac{x+2}{x-1} + \dfrac{x-1}{x+2} = \dfrac{5}{2}$

Javoblar va yechim yo'llari

1. $t = x^2 \ge 0 \implies t^2 - 10t + 9 = 0 \implies t_1 = 1, t_2 = 9$. $x = \pm 1, \pm 3$. Javob: $x = \pm 1,\; \pm 3$.
2. $t = x^2 \ge 0 \implies t^2 - 8t - 9 = 0 \implies t_1 = 9, t_2 = -1$. $x^2 = 9 \implies x = \pm 3$ ($t = -1$ da ildiz yo'q). Javob: $x = \pm 3$.
3. $t = x^2 \ge 0 \implies t^2 = 16 \implies t = 4$. $x^2 = 4 \implies x = \pm 2$. Javob: $x = \pm 2$.
4. $t = x^2 + 2x \implies t^2 - 2t - 3 = 0 \implies t_1 = 3, t_2 = -1$. $x^2 + 2x - 3 = 0 \implies x = 1, -3$; $x^2 + 2x + 1 = 0 \implies x = -1$. Javob: $x_1 = -3,\; x_2 = -1,\; x_3 = 1$.
5. Aniqlanish sohasi $x \neq 4$. Surat $x(x - 4) = 0 \implies x = 0$ yoki $x = 4$ (chet ildiz). Javob: $x = 0$.
6. Aniqlanish sohasi $x \neq \pm 3$. Umumiy maxrajga ko'paytirsak: $x(x+3) - 3(x-3) = 18 \implies x^2 + 9 = 18 \implies x^2 = 9 \implies x = \pm 3$. Ikkala qiymat ham maxrajni nol qiladi. Javob: Ildizga ega emas.
7. $x \neq 1$. $x^2 = 4x - 3 \implies x^2 - 4x + 3 = 0 \implies x_1 = 3, x_2 = 1$ (chet ildiz). Javob: $x = 3$.
8. $t = x^2 \ge 0 \implies t^2 + 5t + 6 = 0 \implies t_1 = -2, t_2 = -3$. Ikkala $t$ ham manfiy, haqiqiy ildiz bermaydi. Javob: Haqiqiy ildiz yo'q.
9. $t = \dfrac{x-1}{x} \implies t^2 - 3t + 2 = 0 \implies t_1 = 1, t_2 = 2$. $\dfrac{x-1}{x} = 1 \implies x - 1 = x \implies -1 = 0$ (ildiz yo'q); $\dfrac{x-1}{x} = 2 \implies x - 1 = 2x \implies x = -1$. Javob: $x = -1$.
10. $t = \dfrac{x+2}{x-1} \implies t + \dfrac{1}{t} = \dfrac{5}{2} \implies 2t^2 - 5t + 2 = 0 \implies t_1 = 2, t_2 = \dfrac{1}{2}$. $\dfrac{x+2}{x-1} = 2 \implies x + 2 = 2x - 2 \implies x = 4$; $\dfrac{x+2}{x-1} = \dfrac{1}{2} \implies 2x + 4 = x - 1 \implies x = -5$. Javob: $x_1 = -5,\; x_2 = 4$.

Keng tarqalgan xatolar

1. Bikvadrat tenglamada manfiy $t$ dan ildiz chiqarish. ✗ $t_2 = -9 \implies x = \pm 3$ ✓ Haqiqiy sonning kvadrati manfiy bo'lolmaydi: $x^2 = -9$ tenglama haqiqiy ildizga ega emas.

2. $t$ ni topib, $x$ ga qaytishni unutish. ✗ $x^4 - 5x^2 + 4 = 0$ da javobni $t_1 = 1, t_2 = 4$ deb qoldirish ✓ Boshlang'ich noma'lum $x$ edi: $x^2 = 1 \implies x = \pm 1$ va $x^2 = 4 \implies x = \pm 2$.

3. Kasr tenglamalarda chet ildizlarni tekshirmaslik. ✗ Maxrajni nolga aylantiradigan qiymatni ham javobga kiritish ✓ Har doim topilgan ildizlarni aniqlanish sohasi bilan solishtiring: maxraj nolga teng bo'lmasligi shart!

4. O'zgaruvchi kiritishda barcha hadlarni almashtirmaslik. ✗ $(x^2 - 2x)^2 - 2x^2 + 4x - 3 = 0$ da faqat bir qismini $t$ deb olish ✓ Guruhlang: $(x^2 - 2x)^2 - 2(x^2 - 2x) - 3 = 0 \implies t^2 - 2t - 3 = 0$.

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Keyingi qadam

Kvadrat tenglamaga keltiriladigan turli tenglamalarni o'rgandik. Endi matematikadagi eng muhim amaliy bosqichga o'tamiz: hayotiy, geometrik va harakatga oid masalalarni kvadrat tenglama tuzib yechish — Kvadrat tenglama yordamida masalalar yechish.