find value of x:

under root of 4x-9 +
under root of 4x+9 = 5+ root7

in this question in the last step we equalize 8x=32 with what concept?

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \sqrt{4\mathrm{x}-9}+\sqrt{4\mathrm{x}+9}=5+\sqrt{7} ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{As} \mathrm{quantity} \mathrm{inside} \mathrm{root} \mathrm{should} \mathrm{be} \mathrm{positive}, \mathrm{hence} \\ 4\mathrm{x}-9\ge 0 \mathrm{and} 4\mathrm{x}+9\ge 0 \\ \Rightarrow \mathrm{x}\ge \frac{9}{4} \mathrm{and} \mathrm{x}\ge -\frac{9}{4} \\ \mathrm{Taking} \mathrm{intersection} \mathrm{of} \mathrm{two} \mathrm{conditions} \mathrm{we} \mathrm{get}, \\ \mathbf{x}\mathbf{\ge }\frac{\mathbf{9}}{\mathbf{4}} \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}, \\ {\left(\sqrt{4\mathrm{x}-9}+\sqrt{4\mathrm{x}+9}\right)}^2={\left(5+\sqrt{7}\right)}^2 \\ \mathrm{Using} \mathrm{idenitiy} {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}, \mathrm{we} \mathrm{get}, \\ \Rightarrow 4\mathrm{x}-9+4\mathrm{x}+9+2\sqrt{\left(4\mathrm{x}-9\right)\left(4\mathrm{x}+9\right)}=25+7+10\sqrt{7} \\ \Rightarrow 8\mathrm{x}+2\sqrt{16{\mathrm{x}}^2-81}=32+10\sqrt{7} \\ \mathrm{Dividing} \mathrm{each} \mathrm{term} \mathrm{by} \text{'}2\text{'} \mathrm{we} \mathrm{get}, \\ \Rightarrow 4\mathrm{x}+\sqrt{16{\mathrm{x}}^2-81}=16+5\sqrt{7} \\ \Rightarrow \sqrt{16{\mathrm{x}}^2-81}=\left\{\left(16+5\sqrt{7}\right)-4\mathrm{x}\right\} \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}, \\ 16{\mathrm{x}}^2-81={\left\{\left(16+5\sqrt{7}\right)-4\mathrm{x}\right\}}^2 \\ \Rightarrow 16{\mathrm{x}}^2-81={\left(16+5\sqrt{7}\right)}^2+16{\mathrm{x}}^2-2.4\mathrm{x}.\left(16+5\sqrt{7}\right) \\ \Rightarrow -81=256+175+160\sqrt{7}-8\mathrm{x}\left(16+5\sqrt{7}\right) \\ \Rightarrow 8\mathrm{x}\left(16+5\sqrt{7}\right)=256+175+81+160\sqrt{7} \\ \Rightarrow 8\mathrm{x}\left(16+5\sqrt{7}\right)=512+160\sqrt{7} \\ \Rightarrow 8\mathrm{x}\left(16+5\sqrt{7}\right)=32\left(16+5\sqrt{7}\right) \\ \Rightarrow 8\mathrm{x}=32 \\ \Rightarrow \mathrm{x}=4 \mathrm{which} \mathrm{satisfies} \mathrm{the} \mathrm{constraint} \mathrm{x}\ge \frac{9}{4}. \\ \mathbf{Hence}\mathbf{ }\mathbf{solution}\mathbf{ }\mathbf{of}\mathbf{ }\sqrt{\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{9}}\mathbf{+}\sqrt{\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{9}}\mathbf{=}\mathbf{5}\mathbf{+}\sqrt{\mathbf{7}}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{.} \end{gathered}$$

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