-4/3x=1/2x^2 eching

$$- \frac{ 4 }{ 3 } x= \frac{ 1 }{ 2 } { x }^{ 2 }$$

x uchun yechish

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2,666666667$$

$$x=0$$

Faktorlash qadamlari

Ikkala tarafdan \frac{1}{2}x^{2} ni ayirish.

$$-\frac{4}{3}x-\frac{1}{2}x^{2}=0$$

x omili.

$$x\left(-\frac{4}{3}-\frac{1}{2}x\right)=0$$

Tenglamani yechish uchun x=0 va -\frac{4}{3}-\frac{x}{2}=0 ni yeching.

$$x=0 x=-\frac{8}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan \frac{1}{2}x^{2} ni ayirish.

$$-\frac{4}{3}x-\frac{1}{2}x^{2}=0$$

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.

$$-\frac{1}{2}x^{2}-\frac{4}{3}x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -\frac{1}{2} ni a, -\frac{4}{3} ni b va 0 ni c bilan almashtiring.

$$x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}}}{2\left(-\frac{1}{2}\right)}$$

\left(-\frac{4}{3}\right)^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-\frac{4}{3}\right)±\frac{4}{3}}{2\left(-\frac{1}{2}\right)}$$

-\frac{4}{3} ning teskarisi \frac{4}{3} ga teng.

$$x=\frac{\frac{4}{3}±\frac{4}{3}}{2\left(-\frac{1}{2}\right)}$$

2 ni -\frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{\frac{4}{3}±\frac{4}{3}}{-1}$$

x=\frac{\frac{4}{3}±\frac{4}{3}}{-1} tenglamasini yeching, bunda ± musbat. Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{4}{3} ni \frac{4}{3} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{\frac{8}{3}}{-1}$$

\frac{8}{3} ni -1 ga bo'lish.

$$x=-\frac{8}{3}$$

x=\frac{\frac{4}{3}±\frac{4}{3}}{-1} tenglamasini yeching, bunda ± manfiy. Umumiy maxrajni topib va suratlarni ayirib \frac{4}{3} ni \frac{4}{3} dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$x=\frac{0}{-1}$$

0 ni -1 ga bo'lish.

$$x=0$$

Tenglama yechildi.

$$x=-\frac{8}{3} x=0$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan \frac{1}{2}x^{2} ni ayirish.

$$-\frac{4}{3}x-\frac{1}{2}x^{2}=0$$

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

$$-\frac{1}{2}x^{2}-\frac{4}{3}x=0$$

Ikkala tarafini -2 ga ko‘paytiring.

$$\frac{-\frac{1}{2}x^{2}-\frac{4}{3}x}{-\frac{1}{2}}=\frac{0}{-\frac{1}{2}}$$

-\frac{1}{2} ga bo'lish -\frac{1}{2} ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{\frac{4}{3}}{-\frac{1}{2}}\right)x=\frac{0}{-\frac{1}{2}}$$

-\frac{4}{3} ni -\frac{1}{2} ga bo'lish -\frac{4}{3} ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x^{2}+\frac{8}{3}x=\frac{0}{-\frac{1}{2}}$$

0 ni -\frac{1}{2} ga bo'lish 0 ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x^{2}+\frac{8}{3}x=0$$

\frac{8}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{4}{3} olish uchun. Keyin, \frac{4}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{4}{3} kvadratini chiqarish.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{16}{9}$$

x^{2}+\frac{8}{3}x+\frac{16}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{16}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Qisqartirish.

$$x+\frac{4}{3}=\frac{4}{3} x+\frac{4}{3}=-\frac{4}{3}$$

Tenglamaning ikkala tarafidan \frac{4}{3} ni ayirish.

$$x=0 x=-\frac{8}{3}$$