Riješi f(x)=8-b*sqrt{-x^2+6x+16}=pi

$$f ( x ) = 8 - b \cdot \sqrt { - x ^ { 2 } + 6 x + 16 } = \pi$$

Izračunaj b (complex solution)

$$b=\left(8-\pi \right)\left(-\left(x-8\right)\left(x+2\right)\right)^{-\frac{1}{2}}$$

$$x\neq -2\text{ and }x\neq 8$$

Koraci rješavanja linearne jednadžbe

Oduzmite 8 od obiju strana.

$$-b\sqrt{-x^{2}+6x+16}=\pi -8$$

Jednadžba je u standardnom obliku.

$$\left(-\sqrt{16+6x-x^{2}}\right)b=\pi -8$$

Podijelite obje strane sa -\sqrt{-x^{2}+6x+16}.

$$\frac{\left(-\sqrt{16+6x-x^{2}}\right)b}{-\sqrt{16+6x-x^{2}}}=\frac{\pi -8}{-\sqrt{16+6x-x^{2}}}$$

Dijeljenjem s -\sqrt{-x^{2}+6x+16} poništava se množenje s -\sqrt{-x^{2}+6x+16}.

$$b=\frac{\pi -8}{-\sqrt{16+6x-x^{2}}}$$

Podijelite \pi -8 s -\sqrt{-x^{2}+6x+16}.

$$b=-\left(\pi -8\right)\left(-\left(x-8\right)\left(x+2\right)\right)^{-\frac{1}{2}}$$

Izračunaj b

$$b=\frac{8-\pi }{\sqrt{\left(8-x\right)\left(x+2\right)}}$$

$$x>-2\text{ and }x<8$$

Koraci rješavanja linearne jednadžbe

Oduzmite 8 od obiju strana.

$$-b\sqrt{-x^{2}+6x+16}=\pi -8$$

Jednadžba je u standardnom obliku.

$$\left(-\sqrt{16+6x-x^{2}}\right)b=\pi -8$$

Podijelite obje strane sa -\sqrt{-x^{2}+6x+16}.

$$\frac{\left(-\sqrt{16+6x-x^{2}}\right)b}{-\sqrt{16+6x-x^{2}}}=\frac{\pi -8}{-\sqrt{16+6x-x^{2}}}$$

Dijeljenjem s -\sqrt{-x^{2}+6x+16} poništava se množenje s -\sqrt{-x^{2}+6x+16}.

$$b=\frac{\pi -8}{-\sqrt{16+6x-x^{2}}}$$

Podijelite \pi -8 s -\sqrt{-x^{2}+6x+16}.

$$b=-\frac{\pi -8}{\sqrt{\left(8-x\right)\left(x+2\right)}}$$

Izračunaj x (complex solution)

$$x=-\frac{\sqrt{25b^{4}-\left(\pi b-8b\right)^{2}}}{b^{2}}+3$$

$$x=\frac{\sqrt{25b^{4}-\left(\pi b-8b\right)^{2}}}{b^{2}}+3\text{, }arg(\frac{1}{b})<\pi \text{ and }b\neq 0$$

Izračunaj x

$$x=-\frac{\sqrt{25b^{2}-\pi ^{2}+16\pi -64}}{b}+3$$

$$x=\frac{\sqrt{25b^{2}-\pi ^{2}+16\pi -64}}{b}+3\text{, }b\geq \frac{8-\pi }{5}$$