a^2(b^2+1)+b(b+2a)=40 ਨੂੰ ਹੱਲ ਕਰੋ

$$a ^ { 2 } ( b ^ { 2 } + 1 ) + b ( b + 2 a ) = 40$$

a ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}a=\frac{\sqrt{40+40b^{2}-b^{4}}-b}{b^{2}+1}\text{; }a=-\frac{\sqrt{40+40b^{2}-b^{4}}+b}{b^{2}+1}\text{, }&b\neq -i\text{ and }b\neq i\\a=\frac{40-b^{2}}{2b}\text{, }&b=-i\text{ or }b=i\end{matrix}\right.$$

b ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}b=\frac{\sqrt{40+40a^{2}-a^{4}}-a}{a^{2}+1}\text{; }b=-\frac{\sqrt{40+40a^{2}-a^{4}}+a}{a^{2}+1}\text{, }&a\neq -i\text{ and }a\neq i\\b=\frac{40-a^{2}}{2a}\text{, }&a=-i\text{ or }a=i\end{matrix}\right.$$

a ਲਈ ਹਲ ਕਰੋ

$$a=\frac{\sqrt{40+40b^{2}-b^{4}}-b}{b^{2}+1}$$

$$a=-\frac{\sqrt{40+40b^{2}-b^{4}}+b}{b^{2}+1}\text{, }|b|\leq \sqrt{2\sqrt{110}+20}$$

b ਲਈ ਹਲ ਕਰੋ

$$b=\frac{\sqrt{40+40a^{2}-a^{4}}-a}{a^{2}+1}$$

$$b=-\frac{\sqrt{40+40a^{2}-a^{4}}+a}{a^{2}+1}\text{, }|a|\leq \sqrt{2\sqrt{110}+20}$$