left(5xright)^2-4x-5=0 eching

$${ \left(5x \right) }^{ 2 } -4x-5 = 0$$

x uchun yechish

$$x=\frac{\sqrt{129}+2}{25}\approx 0,534312668$$

$$x=\frac{2-\sqrt{129}}{25}\approx -0,374312668$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(5x\right)^{2} ni kengaytirish.

$$5^{2}x^{2}-4x-5=0$$

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25x^{2}-4x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, -4 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 25\left(-5\right)}}{2\times 25}$$

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 25\left(-5\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16-100\left(-5\right)}}{2\times 25}$$

-100 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+500}}{2\times 25}$$

16 ni 500 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{516}}{2\times 25}$$

516 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±2\sqrt{129}}{2\times 25}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±2\sqrt{129}}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{4±2\sqrt{129}}{50}$$

x=\frac{4±2\sqrt{129}}{50} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{129} ga qo'shish.

$$x=\frac{2\sqrt{129}+4}{50}$$

4+2\sqrt{129} ni 50 ga bo'lish.

$$x=\frac{\sqrt{129}+2}{25}$$

x=\frac{4±2\sqrt{129}}{50} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{129} ni ayirish.

$$x=\frac{4-2\sqrt{129}}{50}$$

4-2\sqrt{129} ni 50 ga bo'lish.

$$x=\frac{2-\sqrt{129}}{25}$$

Tenglama yechildi.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(5x\right)^{2} ni kengaytirish.

$$5^{2}x^{2}-4x-5=0$$

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25x^{2}-4x-5=0$$

5 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$25x^{2}-4x=5$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}-4x}{25}=\frac{5}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{4}{25}x=\frac{5}{25}$$

\frac{5}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{4}{25}x=\frac{1}{5}$$

-\frac{4}{25} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{2}{25} olish uchun. Keyin, -\frac{2}{25} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{4}{25}x+\left(-\frac{2}{25}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{25}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{25} kvadratini chiqarish.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{1}{5}+\frac{4}{625}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{5} ni \frac{4}{625} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{129}{625}$$

x^{2}-\frac{4}{25}x+\frac{4}{625} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{2}{25}\right)^{2}=\frac{129}{625}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{2}{25}\right)^{2}}=\sqrt{\frac{129}{625}}$$

Qisqartirish.

$$x-\frac{2}{25}=\frac{\sqrt{129}}{25} x-\frac{2}{25}=-\frac{\sqrt{129}}{25}$$

\frac{2}{25} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$