2x^2-15x-1=0 eching

$$2 x ^ { 2 } - 15 x - 1 = 0$$

x uchun yechish

$$x = \frac{\sqrt{233} + 15}{4} \approx 7,566084381$$

$$x=\frac{15-\sqrt{233}}{4}\approx -0,066084381$$

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.

$$2x^{2}-15x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -15 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

-15 kvadratini chiqarish.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\left(-1\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-15\right)±\sqrt{225+8}}{2\times 2}$$

225 ni 8 ga qo'shish.

$$x=\frac{-\left(-15\right)±\sqrt{233}}{2\times 2}$$

-15 ning teskarisi 15 ga teng.

$$x=\frac{15±\sqrt{233}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{15±\sqrt{233}}{4}$$

x=\frac{15±\sqrt{233}}{4} tenglamasini yeching, bunda ± musbat. 15 ni \sqrt{233} ga qo'shish.

$$x=\frac{\sqrt{233}+15}{4}$$

x=\frac{15±\sqrt{233}}{4} tenglamasini yeching, bunda ± manfiy. 15 dan \sqrt{233} ni ayirish.

$$x=\frac{15-\sqrt{233}}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{233}+15}{4} x=\frac{15-\sqrt{233}}{4}$$

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.

$$2x^{2}-15x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}-15x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$2x^{2}-15x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$2x^{2}-15x=1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-15x}{2}=\frac{1}{2}$$

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

$$x^{2}-\frac{15}{2}x=\frac{1}{2}$$

-\frac{15}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{15}{4} olish uchun. Keyin, -\frac{15}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{15}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{15}{4} kvadratini chiqarish.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{1}{2}+\frac{225}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{225}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{233}{16}$$

x^{2}-\frac{15}{2}x+\frac{225}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{233}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{233}{16}}$$

Qisqartirish.

$$x-\frac{15}{4}=\frac{\sqrt{233}}{4} x-\frac{15}{4}=-\frac{\sqrt{233}}{4}$$

\frac{15}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{233}+15}{4} x=\frac{15-\sqrt{233}}{4}$$