(10-2x)x=14 eching

$$(10-2x)x=14$$

x uchun yechish (complex solution)

$$x=\frac{5+\sqrt{3}i}{2}\approx 2,5+0,866025404i$$

$$x=\frac{-\sqrt{3}i+5}{2}\approx 2,5-0,866025404i$$

Kvadrat tenglamalaridan foydalanish qoidalari

10-2x ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x-2x^{2}=14$$

Ikkala tarafdan 14 ni ayirish.

$$10x-2x^{2}-14=0$$

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.

$$-2x^{2}+10x-14=0$$

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 -14 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+8\left(-14\right)}}{2\left(-2\right)}$$

8 ni -14 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-112}}{2\left(-2\right)}$$

100 ni -112 ga qo'shish.

$$x=\frac{-10±\sqrt{-12}}{2\left(-2\right)}$$

-12 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±2\sqrt{3}i}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{-10±2\sqrt{3}i}{-4}$$

x=\frac{-10±2\sqrt{3}i}{-4} tenglamasini yeching, bunda ± musbat. -10 ni 2i\sqrt{3} ga qo'shish.

$$x=\frac{-10+2\sqrt{3}i}{-4}$$

-10+2i\sqrt{3} ni -4 ga bo'lish.

$$x=\frac{-\sqrt{3}i+5}{2}$$

x=\frac{-10±2\sqrt{3}i}{-4} tenglamasini yeching, bunda ± manfiy. -10 dan 2i\sqrt{3} ni ayirish.

$$x=\frac{-2\sqrt{3}i-10}{-4}$$

-10-2i\sqrt{3} ni -4 ga bo'lish.

$$x=\frac{5+\sqrt{3}i}{2}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{3}i+5}{2} x=\frac{5+\sqrt{3}i}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

10-2x ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x-2x^{2}=14$$

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

$$-2x^{2}+10x=14$$

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}+10x}{-2}=\frac{14}{-2}$$

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

$$x^{2}+\frac{10}{-2}x=\frac{14}{-2}$$

10 ni -2 ga bo'lish.

$$x^{2}-5x=\frac{14}{-2}$$

14 ni -2 ga bo'lish.

$$x^{2}-5x=-7$$

-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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-7+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-7+\frac{25}{4}$$

-7 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=-\frac{3}{4}$$

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.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{3}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{3}i}{2} x-\frac{5}{2}=-\frac{\sqrt{3}i}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i+5}{2}$$