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Question Number 186399 by normans last updated on 04/Feb/23

Answered by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

$$a+b_1=b_2+c$$$$\Rightarrow b_1=c-a+b_2$$$$\frac{b_2}{c}=\frac{h}{p}$$$$\frac{b_1}{a}=\frac{h}{p-2h}=\frac{1}{\frac{p}{h}-2}=\frac{1}{\frac{c}{b_2}-2}=\frac{b_2}{c-2b_2}$$$$\frac{c-a+b_2}{a}=\frac{b_2}{c-2b_2}$$$$2b_2^2+(c-a)b_2-c(c-a)=0$$$$b_2=\frac{-(c-a)+\sqrt{(c-a)(9c-a)}}{4}$$$$b=b_1+b_2=c-a+2b_2$$$$\Rightarrow b=\frac{c-a+\sqrt{(c-a)(9c-a)}}{2}✓$$$$x=\frac{28-27+\sqrt{(28-27)(9\times 28-27)}}{2}=8✓$$

Commented by normans last updated on 04/Feb/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{h}\mathit{o}\mathit{u},\mathit{g}\mathit{r}\mathit{e}\mathit{a}\mathit{t}\mathit{S}ir$$

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