5x^2+25x-10=0 eching

$$5 x ^ { 2 } + 25 x - 10 = 0$$

x uchun yechish

$$x=\frac{\sqrt{33}-5}{2}\approx 0,372281323$$

$$x=\frac{-\sqrt{33}-5}{2}\approx -5,372281323$$

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.

$$5x^{2}+25x-10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 25 ni b va -10 ni c bilan almashtiring.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5\left(-10\right)}}{2\times 5}$$

25 kvadratini chiqarish.

$$x=\frac{-25±\sqrt{625-4\times 5\left(-10\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-20\left(-10\right)}}{2\times 5}$$

-20 ni -10 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625+200}}{2\times 5}$$

625 ni 200 ga qo'shish.

$$x=\frac{-25±\sqrt{825}}{2\times 5}$$

825 ning kvadrat ildizini chiqarish.

$$x=\frac{-25±5\sqrt{33}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-25±5\sqrt{33}}{10}$$

x=\frac{-25±5\sqrt{33}}{10} tenglamasini yeching, bunda ± musbat. -25 ni 5\sqrt{33} ga qo'shish.

$$x=\frac{5\sqrt{33}-25}{10}$$

-25+5\sqrt{33} ni 10 ga bo'lish.

$$x=\frac{\sqrt{33}-5}{2}$$

x=\frac{-25±5\sqrt{33}}{10} tenglamasini yeching, bunda ± manfiy. -25 dan 5\sqrt{33} ni ayirish.

$$x=\frac{-5\sqrt{33}-25}{10}$$

-25-5\sqrt{33} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{33}-5}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{33}-5}{2} x=\frac{-\sqrt{33}-5}{2}$$

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.

$$5x^{2}+25x-10=0$$

10 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+25x-10-\left(-10\right)=-\left(-10\right)$$

O‘zidan -10 ayirilsa 0 qoladi.

$$5x^{2}+25x=-\left(-10\right)$$

0 dan -10 ni ayirish.

$$5x^{2}+25x=10$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+25x}{5}=\frac{10}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{25}{5}x=\frac{10}{5}$$

25 ni 5 ga bo'lish.

$$x^{2}+5x=\frac{10}{5}$$

10 ni 5 ga bo'lish.

$$x^{2}+5x=2$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=2+\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

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

2 ni \frac{25}{4} ga qo'shish.

$$x^{2}+5x+\frac{25}{4}=\frac{33}{4}$$

x^{2}+5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{33}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Qisqartirish.

$$x+\frac{5}{2}=\frac{\sqrt{33}}{2} x+\frac{5}{2}=-\frac{\sqrt{33}}{2}$$

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

$$x=\frac{\sqrt{33}-5}{2} x=\frac{-\sqrt{33}-5}{2}$$