x^2-5x-130=0 eching

$$x ^ { 2 } - 5 x - 130 = 0$$

x uchun yechish

$$x = \frac{\sqrt{545} + 5}{2} \approx 14,17261753$$

$$x=\frac{5-\sqrt{545}}{2}\approx -9,17261753$$

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.

$$x^{2}-5x-130=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -5 ni b va -130 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-130\right)}}{2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-130\right)}}{2}$$

-4 ni -130 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+520}}{2}$$

25 ni 520 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{545}}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{545}}{2}$$

x=\frac{5±\sqrt{545}}{2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{545} ga qo'shish.

$$x=\frac{\sqrt{545}+5}{2}$$

x=\frac{5±\sqrt{545}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{545} ni ayirish.

$$x=\frac{5-\sqrt{545}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{545}+5}{2} x=\frac{5-\sqrt{545}}{2}$$

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.

$$x^{2}-5x-130=0$$

130 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-5x-130-\left(-130\right)=-\left(-130\right)$$

O‘zidan -130 ayirilsa 0 qoladi.

$$x^{2}-5x=-\left(-130\right)$$

0 dan -130 ni ayirish.

$$x^{2}-5x=130$$

-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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=130+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=130+\frac{25}{4}$$

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

$$x^{2}-5x+\frac{25}{4}=\frac{545}{4}$$

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.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{545}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{545}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{545}}{2} x-\frac{5}{2}=-\frac{\sqrt{545}}{2}$$

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

$$x=\frac{\sqrt{545}+5}{2} x=\frac{5-\sqrt{545}}{2}$$