x^2+3x-65=10 eching

$${ x }^{ 2 } +3x-65=10$$

x uchun yechish

$$x = \frac{\sqrt{309} - 3}{2} \approx 7,289197916$$

$$x=\frac{-\sqrt{309}-3}{2}\approx -10,289197916$$

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.

$$x^{2}+3x-65=10$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$x^{2}+3x-65-10=10-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$x^{2}+3x-65-10=0$$

-65 dan 10 ni ayirish.

$$x^{2}+3x-75=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 3 ni b va -75 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-75\right)}}{2}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-75\right)}}{2}$$

-4 ni -75 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+300}}{2}$$

9 ni 300 ga qo'shish.

$$x=\frac{-3±\sqrt{309}}{2}$$

x=\frac{-3±\sqrt{309}}{2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{309} ga qo'shish.

$$x=\frac{\sqrt{309}-3}{2}$$

x=\frac{-3±\sqrt{309}}{2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{309} ni ayirish.

$$x=\frac{-\sqrt{309}-3}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{309}-3}{2} x=\frac{-\sqrt{309}-3}{2}$$

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.

$$x^{2}+3x-65=10$$

65 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+3x-65-\left(-65\right)=10-\left(-65\right)$$

O‘zidan -65 ayirilsa 0 qoladi.

$$x^{2}+3x=10-\left(-65\right)$$

10 dan -65 ni ayirish.

$$x^{2}+3x=75$$

3 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{2} olish uchun. Keyin, \frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=75+\left(\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=75+\frac{9}{4}$$

75 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=\frac{309}{4}$$

x^{2}+3x+\frac{9}{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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{309}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{309}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{309}}{2} x+\frac{3}{2}=-\frac{\sqrt{309}}{2}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$x=\frac{\sqrt{309}-3}{2} x=\frac{-\sqrt{309}-3}{2}$$