x^2+15x-2=0 eching

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

x uchun yechish

$$x=\frac{\sqrt{233}-15}{2}\approx 0,132168761$$

$$x=\frac{-\sqrt{233}-15}{2}\approx -15,132168761$$

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.

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

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

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

15 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{225+8}}{2}$$

225 ni 8 ga qo'shish.

$$x=\frac{-15±\sqrt{233}}{2}$$

x=\frac{-15±\sqrt{233}}{2} tenglamasini yeching, bunda ± musbat. -15 ni \sqrt{233} ga qo'shish.

$$x=\frac{\sqrt{233}-15}{2}$$

x=\frac{-15±\sqrt{233}}{2} tenglamasini yeching, bunda ± manfiy. -15 dan \sqrt{233} ni ayirish.

$$x=\frac{-\sqrt{233}-15}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{233}-15}{2} x=\frac{-\sqrt{233}-15}{2}$$

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.

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

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

0 dan -2 ni ayirish.

$$x^{2}+15x=2$$

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{15}{2} kvadratini chiqarish.

$$x^{2}+15x+\frac{225}{4}=2+\frac{225}{4}$$

2 ni \frac{225}{4} ga qo'shish.

$$x^{2}+15x+\frac{225}{4}=\frac{233}{4}$$

x^{2}+15x+\frac{225}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+\frac{15}{2}=\frac{\sqrt{233}}{2} x+\frac{15}{2}=-\frac{\sqrt{233}}{2}$$

Tenglamaning ikkala tarafidan \frac{15}{2} ni ayirish.

$$x=\frac{\sqrt{233}-15}{2} x=\frac{-\sqrt{233}-15}{2}$$