(sqrt{a}+sqrt{8a})^2=54+bsqrt{2} eching

$$( \sqrt { a } + \sqrt { 8 a } ) ^ { 2 } = 54 + b \sqrt { 2 }$$

b uchun yechish

$$b=\frac{9\sqrt{2}a}{2}+4a-27\sqrt{2}$$

$$a\geq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(p+q\right)^{2}=p^{2}+2pq+q^{2} binom teoremasini \left(\sqrt{a}+\sqrt{8a}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{a}\right)^{2}+2\sqrt{a}\sqrt{8a}+\left(\sqrt{8a}\right)^{2}=54+b\sqrt{2}$$

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

$$a+2\sqrt{a}\sqrt{8a}+\left(\sqrt{8a}\right)^{2}=54+b\sqrt{2}$$

2 daraja ko‘rsatkichini \sqrt{8a} ga hisoblang va 8a ni qiymatni oling.

$$a+2\sqrt{a}\sqrt{8a}+8a=54+b\sqrt{2}$$

9a ni olish uchun a va 8a ni birlashtirish.

$$9a+2\sqrt{a}\sqrt{8a}=54+b\sqrt{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$54+b\sqrt{2}=9a+2\sqrt{a}\sqrt{8a}$$

Ikkala tarafdan 54 ni ayirish.

$$b\sqrt{2}=9a+2\sqrt{a}\sqrt{8a}-54$$

Tenglama standart shaklda.

$$\sqrt{2}b=2\sqrt{a}\sqrt{8a}+9a-54$$

Ikki tarafini \sqrt{2} ga bo‘ling.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{4\sqrt{2}a+9a-54}{\sqrt{2}}$$

\sqrt{2} ga bo'lish \sqrt{2} ga ko'paytirishni bekor qiladi.

$$b=\frac{4\sqrt{2}a+9a-54}{\sqrt{2}}$$

9a+4a\sqrt{2}-54 ni \sqrt{2} ga bo'lish.

$$b=\frac{\sqrt{2}\left(4\sqrt{2}a+9a-54\right)}{2}$$

a uchun yechish

$$a=-\frac{\left(4\sqrt{2}-9\right)\left(\sqrt{2}b+54\right)}{49}$$

$$b\geq -27\sqrt{2}$$