4t^2-10t=0 eching

$$4 t ^ { 2 } - 10 t = 0$$

t uchun yechish

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

$$t=0$$

Faktorlash qadamlari

t omili.

$$t\left(4t-10\right)=0$$

Tenglamani yechish uchun t=0 va 4t-10=0 ni yeching.

$$t=0 t=\frac{5}{2}$$

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.

$$4t^{2}-10t=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, -10 ni b va 0 ni c bilan almashtiring.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 4}$$

\left(-10\right)^{2} ning kvadrat ildizini chiqarish.

$$t=\frac{-\left(-10\right)±10}{2\times 4}$$

-10 ning teskarisi 10 ga teng.

$$t=\frac{10±10}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$t=\frac{10±10}{8}$$

t=\frac{10±10}{8} tenglamasini yeching, bunda ± musbat. 10 ni 10 ga qo'shish.

$$t=\frac{20}{8}$$

\frac{20}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t=\frac{5}{2}$$

t=\frac{10±10}{8} tenglamasini yeching, bunda ± manfiy. 10 dan 10 ni ayirish.

$$t=\frac{0}{8}$$

0 ni 8 ga bo'lish.

$$t=0$$

Tenglama yechildi.

$$t=\frac{5}{2} t=0$$

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.

$$4t^{2}-10t=0$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4t^{2}-10t}{4}=\frac{0}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$t^{2}+\left(-\frac{10}{4}\right)t=\frac{0}{4}$$

\frac{-10}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t^{2}-\frac{5}{2}t=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$t^{2}-\frac{5}{2}t=0$$

-\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{4} olish uchun. Keyin, -\frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$t^{2}-\frac{5}{2}t+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{4} kvadratini chiqarish.

$$t^{2}-\frac{5}{2}t+\frac{25}{16}=\frac{25}{16}$$

t^{2}-\frac{5}{2}t+\frac{25}{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.

$$\left(t-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(t-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Qisqartirish.

$$t-\frac{5}{4}=\frac{5}{4} t-\frac{5}{4}=-\frac{5}{4}$$

\frac{5}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$t=\frac{5}{2} t=0$$