sqrt{2x}+2=1/2x eching

$$\sqrt { 2 x } + 2 = \frac { 1 } { 2 } x$$

x uchun yechish

$$x=4\left(\sqrt{3}+2\right)\approx 14,92820323$$

Yechim qadamlari

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$\sqrt{2x}=\frac{1}{2}x-2$$

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{2x} ga hisoblang va 2x ni qiymatni oling.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\frac{1}{2}x-2\right)^{2} kengaytirilishi uchun ishlating.

$$2x=\frac{1}{4}x^{2}-2x+4$$

Ikkala tarafdan \frac{1}{4}x^{2} ni ayirish.

$$2x-\frac{1}{4}x^{2}=-2x+4$$

2x ni ikki tarafga qo’shing.

$$2x-\frac{1}{4}x^{2}+2x=4$$

4x ni olish uchun 2x va 2x ni birlashtirish.

$$4x-\frac{1}{4}x^{2}=4$$

Ikkala tarafdan 4 ni ayirish.

$$4x-\frac{1}{4}x^{2}-4=0$$

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.

$$-\frac{1}{4}x^{2}+4x-4=0$$

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

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

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{1}{4}\right)\left(-4\right)}}{2\left(-\frac{1}{4}\right)}$$

-4 ni -\frac{1}{4} marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-4}}{2\left(-\frac{1}{4}\right)}$$

16 ni -4 ga qo'shish.

$$x=\frac{-4±\sqrt{12}}{2\left(-\frac{1}{4}\right)}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±2\sqrt{3}}{2\left(-\frac{1}{4}\right)}$$

2 ni -\frac{1}{4} marotabaga ko'paytirish.

$$x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}}$$

x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}-4}{-\frac{1}{2}}$$

-4+2\sqrt{3} ni -\frac{1}{2} ga bo'lish -4+2\sqrt{3} ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=8-4\sqrt{3}$$

x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{3} ni ayirish.

$$x=\frac{-2\sqrt{3}-4}{-\frac{1}{2}}$$

-4-2\sqrt{3} ni -\frac{1}{2} ga bo'lish -4-2\sqrt{3} ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=4\sqrt{3}+8$$

Tenglama yechildi.

$$x=8-4\sqrt{3} x=4\sqrt{3}+8$$

\sqrt{2x}+2=\frac{1}{2}x tenglamasida x uchun 8-4\sqrt{3} ni almashtiring.

$$\sqrt{2\left(8-4\sqrt{3}\right)}+2=\frac{1}{2}\left(8-4\sqrt{3}\right)$$

Qisqartirish. x=8-4\sqrt{3} qiymati bu tenglamani qoniqtirmaydi.

$$2\times 3^{\frac{1}{2}}=4-2\times 3^{\frac{1}{2}}$$

\sqrt{2x}+2=\frac{1}{2}x tenglamasida x uchun 4\sqrt{3}+8 ni almashtiring.

$$\sqrt{2\left(4\sqrt{3}+8\right)}+2=\frac{1}{2}\left(4\sqrt{3}+8\right)$$

Qisqartirish. x=4\sqrt{3}+8 tenglamani qoniqtiradi.

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

\sqrt{2x}=\frac{x}{2}-2 tenglamasi noyob yechimga ega.

$$x=4\sqrt{3}+8$$