2x^2+10x=0 eching

$$2 { x }^{ 2 } +10x=0$$

x uchun yechish

$$x=-5$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(2x+10\right)=0$$

Tenglamani yechish uchun x=0 va 2x+10=0 ni yeching.

$$x=0 x=-5$$

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.

$$2x^{2}+10x=0$$

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

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 2}$$

10^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-10±10}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-10±10}{4}$$

x=\frac{-10±10}{4} tenglamasini yeching, bunda ± musbat. -10 ni 10 ga qo'shish.

$$x=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$x=0$$

x=\frac{-10±10}{4} tenglamasini yeching, bunda ± manfiy. -10 dan 10 ni ayirish.

$$x=-\frac{20}{4}$$

-20 ni 4 ga bo'lish.

$$x=-5$$

Tenglama yechildi.

$$x=0 x=-5$$

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.

$$2x^{2}+10x=0$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+10x}{2}=\frac{0}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{10}{2}x=\frac{0}{2}$$

10 ni 2 ga bo'lish.

$$x^{2}+5x=\frac{0}{2}$$

0 ni 2 ga bo'lish.

$$x^{2}+5x=0$$

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.

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

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

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

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

$$x=0 x=-5$$