ex^2+3x+4=0 eching

$$e x ^ { 2 } + 3 x + 4 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+i\sqrt{16e-9}}{2e}\approx -0,551819162+1,080283934i$$

$$x=-\frac{3+i\sqrt{16e-9}}{2e}\approx -0,551819162-1,080283934i$$

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.

$$ex^{2}+3x+4=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4e\times 4}}{2e}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4e\times 4}}{2e}$$

-4 ni e marotabaga ko'paytirish.

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

-4e ni 4 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-16e}}{2e}$$

9-16e ning kvadrat ildizini chiqarish.

$$x=\frac{-3±i\sqrt{-\left(9-16e\right)}}{2e}$$

x=\frac{-3±i\sqrt{-\left(9-16e\right)}}{2e} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{-\left(9-16e\right)} ga qo'shish.

$$x=\frac{-3+i\sqrt{16e-9}}{2e}$$

x=\frac{-3±i\sqrt{-\left(9-16e\right)}}{2e} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{-\left(9-16e\right)} ni ayirish.

$$x=\frac{-i\sqrt{16e-9}-3}{2e}$$

-3-i\sqrt{-9+16e} ni 2e ga bo'lish.

$$x=-\frac{3+i\sqrt{16e-9}}{2e}$$

Tenglama yechildi.

$$x=\frac{-3+i\sqrt{16e-9}}{2e} x=-\frac{3+i\sqrt{16e-9}}{2e}$$

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.

$$ex^{2}+3x+4=0$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$ex^{2}+3x+4-4=-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$ex^{2}+3x=-4$$

Ikki tarafini e ga bo‘ling.

$$\frac{ex^{2}+3x}{e}=-\frac{4}{e}$$

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

$$x^{2}+\frac{3}{e}x=-\frac{4}{e}$$

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

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

\frac{3}{2e} kvadratini chiqarish.

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

-\frac{4}{e} ni \frac{9}{4e^{2}} ga qo'shish.

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

x^{2}+\frac{3}{e}x+\frac{9}{4e^{2}} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{2e}\right)^{2}=\frac{\frac{9}{4}-4e}{e^{2}}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2e}\right)^{2}}=\sqrt{\frac{\frac{9}{4}-4e}{e^{2}}}$$

Qisqartirish.

$$x+\frac{3}{2e}=\frac{i\sqrt{-\left(9-16e\right)}}{2e} x+\frac{3}{2e}=-\frac{i\sqrt{16e-9}}{2e}$$

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

$$x=\frac{-3+i\sqrt{16e-9}}{2e} x=-\frac{3+i\sqrt{16e-9}}{2e}$$