Atrisiniet -a=sqrt{2a^2+3ab-10}

$$-a = \sqrt{ 2 { a }^{ 2 } +3ab-10 }$$

Atrast b

$$b=-\frac{a}{3}+\frac{10}{3a}$$

$$a<0$$

Vienādojumu sakņu atrisināšanas darbības

Mainiet puses tā, lai visi mainīgie locekļi atrastos pa kreisi.

$$\sqrt{2a^{2}+3ab-10}=-a$$

Kāpiniet kvadrātā abas vienādojuma puses.

$$3ab+2a^{2}-10=a^{2}$$

Atņemiet 2a^{2}-10 no vienādojuma abām pusēm.

$$3ab+2a^{2}-10-\left(2a^{2}-10\right)=a^{2}-\left(2a^{2}-10\right)$$

Atņemot 2a^{2}-10 no sevis, paliek 0.

$$3ab=a^{2}-\left(2a^{2}-10\right)$$

Atņemiet 2a^{2}-10 no a^{2}.

$$3ab=10-a^{2}$$

Daliet abas puses ar 3a.

$$\frac{3ab}{3a}=\frac{10-a^{2}}{3a}$$

Dalīšana ar 3a atsauc reizināšanu ar 3a.

$$b=\frac{10-a^{2}}{3a}$$

Daliet -a^{2}+10 ar 3a.

$$b=-\frac{a}{3}+\frac{10}{3a}$$

Atrast a (complex solution)

$$\left\{\begin{matrix}a=\frac{-\sqrt{9b^{2}+40}-3b}{2}\text{, }&arg(\frac{-\sqrt{9b^{2}+40}-3b}{2})\geq \pi \\a=\frac{\sqrt{9b^{2}+40}-3b}{2}\text{, }&arg(\frac{\sqrt{9b^{2}+40}-3b}{2})\geq \pi \end{matrix}\right,$$

Atrast b (complex solution)

$$b=-\frac{a}{3}+\frac{10}{3a}$$

$$arg(a)\geq \pi \text{ and }a\neq 0$$

Atrast a

$$a=\frac{-\sqrt{9b^{2}+40}-3b}{2}$$

$$\left(\frac{\sqrt{9b^{2}+80}}{4}-\frac{\sqrt{9b^{2}+40}}{2}-\frac{3b}{4}\leq 0\text{ or }-\frac{\sqrt{9b^{2}+40}}{2}-\frac{\sqrt{9b^{2}+80}}{4}-\frac{3b}{4}\geq 0\right)\text{ and }b\leq \frac{\sqrt{9b^{2}+40}}{6}+\frac{3b}{2}$$