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Question Number 165381 by mathlove last updated on 31/Jan/22

Commented by MJS_new last updated on 02/Feb/22

$$\mathrm{we}\mathrm{have}\mathrm{only}\mathrm{one}\mathrm{equation}\mathrm{for}2\mathrm{unknowns}$$$$\Rightarrow \mathrm{no}\mathrm{unique}\mathrm{solution}.$$

Answered by mahdipoor last updated on 01/Feb/22

$${\mathrm{b}}^2+{\mathrm{a}}^2=2\mathrm{a}-2-\frac{1}{{4\mathrm{b}}^2}\Rightarrow$$$$({\mathrm{b}}^2+\frac{1}{{4\mathrm{b}}^2}-1)+({\mathrm{a}}^2-2\mathrm{a}+1)=-2\Rightarrow$$$${(\mathrm{b}-\frac{1}{2\mathrm{b}})}^2+{(\mathrm{a}-1)}^2=-2\Rightarrow impossible$$

Commented by som(math1967) last updated on 01/Feb/22

$$\mathit{h}\mathit{o}\mathit{w}{\mathit{b}}^2+\frac{1}{4{\mathit{b}}^2}-1={(\mathit{b}-\frac{1}{\mathit{b}})}^2\mathit{s}\mathit{i}\mathit{r}?$$

Commented by mahdipoor last updated on 01/Feb/22

$$thanksforyournot$$

Answered by som(math1967) last updated on 01/Feb/22

$$a^2-2\mathit{a}+{\mathit{b}}^2=-2-\frac{1}{4{\mathit{b}}^2}$$$$\Rightarrow {\mathit{a}}^2-2\mathit{a}+1+{\mathit{b}}^2+{\left(\frac{1}{2\mathit{b}}\right)}^2+2.\mathit{b}.\frac{1}{2\mathit{b}}=0$$$$\Rightarrow {(\mathit{a}-1)}^2+{(\mathit{b}+\frac{1}{2\mathit{b}})}^2=0$$$$\Rightarrow {(\mathit{a}-1)}^2=0\Rightarrow \mathit{a}=1$$$${(\mathit{b}+\frac{1}{2\mathit{b}})}^2=0\Rightarrow \mathit{b}=-\frac{1}{2\mathit{b}}$$$$\therefore 2{\mathit{b}}^2=-1\therefore 6{\mathit{b}}^2=-3$$$$\mathit{s}\mathit{o}6{\mathit{b}}^2+5\mathit{a}=-3+5=2$$

Commented by mahdipoor last updated on 01/Feb/22

$$2b^2=-1\Rightarrow b=\sqrt{-0.5}=\frac{\mathrm{i}}{\sqrt{2}}$$$$bisdefinedinrealnumbers$$$$ifwesay2b^2=-1,alsowecansay:$$$${(a-1)}^2+{(b+\frac{1}{2b})}^2=0\Rightarrow \{\begin{array}{l}{(a-1)}^2=\frac{1}{4}\\ {(b+\frac{1}{2b})}^2=-\frac{1}{4}\end{array}$$$$\Rightarrow a=1\pm \frac{1}{2}\mathrm{\&}b=\pm \mathrm{i}\Rightarrow 6b^2+5a=-1\pm \frac{5}{2}$$

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