-2/3t^2+3t=3 eching

$$- \frac { 2 } { 3 } t ^ { 2 } + 3 t = 3$$

t uchun yechish

$$t=3$$

$$t = \frac{3}{2} = 1\frac{1}{2} = 1,5$$

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.

$$-\frac{2}{3}t^{2}+3t=3$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$-\frac{2}{3}t^{2}+3t-3=3-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$-\frac{2}{3}t^{2}+3t-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -\frac{2}{3} ni a, 3 ni b va -3 ni c bilan almashtiring.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-\frac{2}{3}\right)\left(-3\right)}}{2\left(-\frac{2}{3}\right)}$$

3 kvadratini chiqarish.

$$t=\frac{-3±\sqrt{9-4\left(-\frac{2}{3}\right)\left(-3\right)}}{2\left(-\frac{2}{3}\right)}$$

-4 ni -\frac{2}{3} marotabaga ko'paytirish.

$$t=\frac{-3±\sqrt{9+\frac{8}{3}\left(-3\right)}}{2\left(-\frac{2}{3}\right)}$$

\frac{8}{3} ni -3 marotabaga ko'paytirish.

$$t=\frac{-3±\sqrt{9-8}}{2\left(-\frac{2}{3}\right)}$$

9 ni -8 ga qo'shish.

$$t=\frac{-3±\sqrt{1}}{2\left(-\frac{2}{3}\right)}$$

1 ning kvadrat ildizini chiqarish.

$$t=\frac{-3±1}{2\left(-\frac{2}{3}\right)}$$

2 ni -\frac{2}{3} marotabaga ko'paytirish.

$$t=\frac{-3±1}{-\frac{4}{3}}$$

t=\frac{-3±1}{-\frac{4}{3}} tenglamasini yeching, bunda ± musbat. -3 ni 1 ga qo'shish.

$$t=-\frac{2}{-\frac{4}{3}}$$

-2 ni -\frac{4}{3} ga bo'lish -2 ga k'paytirish -\frac{4}{3} ga qaytarish.

$$t=\frac{3}{2}$$

t=\frac{-3±1}{-\frac{4}{3}} tenglamasini yeching, bunda ± manfiy. -3 dan 1 ni ayirish.

$$t=-\frac{4}{-\frac{4}{3}}$$

-4 ni -\frac{4}{3} ga bo'lish -4 ga k'paytirish -\frac{4}{3} ga qaytarish.

$$t=3$$

Tenglama yechildi.

$$t=\frac{3}{2} t=3$$

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.

$$-\frac{2}{3}t^{2}+3t=3$$

Tenglamaning ikki tarafini -\frac{2}{3} ga bo'lish, bu kasrni qaytarish orqali ikkala tarafga ko'paytirish bilan aynidir.

$$\frac{-\frac{2}{3}t^{2}+3t}{-\frac{2}{3}}=\frac{3}{-\frac{2}{3}}$$

-\frac{2}{3} ga bo'lish -\frac{2}{3} ga ko'paytirishni bekor qiladi.

$$t^{2}+\frac{3}{-\frac{2}{3}}t=\frac{3}{-\frac{2}{3}}$$

3 ni -\frac{2}{3} ga bo'lish 3 ga k'paytirish -\frac{2}{3} ga qaytarish.

$$t^{2}-\frac{9}{2}t=\frac{3}{-\frac{2}{3}}$$

3 ni -\frac{2}{3} ga bo'lish 3 ga k'paytirish -\frac{2}{3} ga qaytarish.

$$t^{2}-\frac{9}{2}t=-\frac{9}{2}$$

-\frac{9}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{9}{4} olish uchun. Keyin, -\frac{9}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$t^{2}-\frac{9}{2}t+\left(-\frac{9}{4}\right)^{2}=-\frac{9}{2}+\left(-\frac{9}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{9}{4} kvadratini chiqarish.

$$t^{2}-\frac{9}{2}t+\frac{81}{16}=-\frac{9}{2}+\frac{81}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{9}{2} ni \frac{81}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$t^{2}-\frac{9}{2}t+\frac{81}{16}=\frac{9}{16}$$

t^{2}-\frac{9}{2}t+\frac{81}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(t-\frac{9}{4}\right)^{2}=\frac{9}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(t-\frac{9}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Qisqartirish.

$$t-\frac{9}{4}=\frac{3}{4} t-\frac{9}{4}=-\frac{3}{4}$$

\frac{9}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$t=3 t=\frac{3}{2}$$