2k^2+6k-2=0 eching

$$2 k ^ { 2 } + 6 k - 2 = 0$$

k uchun yechish

$$k=\frac{\sqrt{13}-3}{2}\approx 0,302775638$$

$$k=\frac{-\sqrt{13}-3}{2}\approx -3,302775638$$

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.

$$2k^{2}+6k-2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 6 ni b va -2 ni c bilan almashtiring.

$$k=\frac{-6±\sqrt{6^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

6 kvadratini chiqarish.

$$k=\frac{-6±\sqrt{36-4\times 2\left(-2\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$k=\frac{-6±\sqrt{36-8\left(-2\right)}}{2\times 2}$$

-8 ni -2 marotabaga ko'paytirish.

$$k=\frac{-6±\sqrt{36+16}}{2\times 2}$$

36 ni 16 ga qo'shish.

$$k=\frac{-6±\sqrt{52}}{2\times 2}$$

52 ning kvadrat ildizini chiqarish.

$$k=\frac{-6±2\sqrt{13}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$k=\frac{-6±2\sqrt{13}}{4}$$

k=\frac{-6±2\sqrt{13}}{4} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{13} ga qo'shish.

$$k=\frac{2\sqrt{13}-6}{4}$$

-6+2\sqrt{13} ni 4 ga bo'lish.

$$k=\frac{\sqrt{13}-3}{2}$$

k=\frac{-6±2\sqrt{13}}{4} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{13} ni ayirish.

$$k=\frac{-2\sqrt{13}-6}{4}$$

-6-2\sqrt{13} ni 4 ga bo'lish.

$$k=\frac{-\sqrt{13}-3}{2}$$

Tenglama yechildi.

$$k=\frac{\sqrt{13}-3}{2} k=\frac{-\sqrt{13}-3}{2}$$

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.

$$2k^{2}+6k-2=0$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$2k^{2}+6k-2-\left(-2\right)=-\left(-2\right)$$

O‘zidan -2 ayirilsa 0 qoladi.

$$2k^{2}+6k=-\left(-2\right)$$

0 dan -2 ni ayirish.

$$2k^{2}+6k=2$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2k^{2}+6k}{2}=\frac{2}{2}$$

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

$$k^{2}+\frac{6}{2}k=\frac{2}{2}$$

6 ni 2 ga bo'lish.

$$k^{2}+3k=\frac{2}{2}$$

2 ni 2 ga bo'lish.

$$k^{2}+3k=1$$

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.

$$k^{2}+3k+\left(\frac{3}{2}\right)^{2}=1+\left(\frac{3}{2}\right)^{2}$$

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

$$k^{2}+3k+\frac{9}{4}=1+\frac{9}{4}$$

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

$$k^{2}+3k+\frac{9}{4}=\frac{13}{4}$$

k^{2}+3k+\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.

$$\left(k+\frac{3}{2}\right)^{2}=\frac{13}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(k+\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Qisqartirish.

$$k+\frac{3}{2}=\frac{\sqrt{13}}{2} k+\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

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

$$k=\frac{\sqrt{13}-3}{2} k=\frac{-\sqrt{13}-3}{2}$$