sqrt{4n+3}=n eching

$$\sqrt{ 4n+3 } = n$$

n uchun yechish

$$n=\sqrt{7}+2\approx 4,645751311$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{4n+3}\right)^{2}=n^{2}$$

2 daraja ko‘rsatkichini \sqrt{4n+3} ga hisoblang va 4n+3 ni qiymatni oling.

$$4n+3=n^{2}$$

Ikkala tarafdan n^{2} ni ayirish.

$$4n+3-n^{2}=0$$

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.

$$-n^{2}+4n+3=0$$

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

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

4 kvadratini chiqarish.

$$n=\frac{-4±\sqrt{16-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$n=\frac{-4±\sqrt{16+4\times 3}}{2\left(-1\right)}$$

4 ni 3 marotabaga ko'paytirish.

$$n=\frac{-4±\sqrt{16+12}}{2\left(-1\right)}$$

16 ni 12 ga qo'shish.

$$n=\frac{-4±\sqrt{28}}{2\left(-1\right)}$$

28 ning kvadrat ildizini chiqarish.

$$n=\frac{-4±2\sqrt{7}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$n=\frac{-4±2\sqrt{7}}{-2}$$

n=\frac{-4±2\sqrt{7}}{-2} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{7} ga qo'shish.

$$n=\frac{2\sqrt{7}-4}{-2}$$

-4+2\sqrt{7} ni -2 ga bo'lish.

$$n=2-\sqrt{7}$$

n=\frac{-4±2\sqrt{7}}{-2} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{7} ni ayirish.

$$n=\frac{-2\sqrt{7}-4}{-2}$$

-4-2\sqrt{7} ni -2 ga bo'lish.

$$n=\sqrt{7}+2$$

Tenglama yechildi.

$$n=2-\sqrt{7} n=\sqrt{7}+2$$

\sqrt{4n+3}=n tenglamasida n uchun 2-\sqrt{7} ni almashtiring.

$$\sqrt{4\left(2-\sqrt{7}\right)+3}=2-\sqrt{7}$$

Qisqartirish. n=2-\sqrt{7} qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$7^{\frac{1}{2}}-2=2-7^{\frac{1}{2}}$$

\sqrt{4n+3}=n tenglamasida n uchun \sqrt{7}+2 ni almashtiring.

$$\sqrt{4\left(\sqrt{7}+2\right)+3}=\sqrt{7}+2$$

Qisqartirish. n=\sqrt{7}+2 tenglamani qoniqtiradi.

$$2+7^{\frac{1}{2}}=2+7^{\frac{1}{2}}$$

\sqrt{4n+3}=n tenglamasi noyob yechimga ega.

$$n=\sqrt{7}+2$$