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Question

Solve the following equation:
$$\displaystyle \sqrt{x \, + \, 1} \, + \, \sqrt{4x \, + \, 13} \, = \, \sqrt{3x \, + \, 12}$$


Solution

$$\sqrt { x+1 } +\sqrt { 4x+13 } =\sqrt { 3x+12 } $$
squaring both sides
$$\Rightarrow x+1+4x+13+2\sqrt { x+1 } \sqrt { 4x+13 } =3x+12$$
$$\Rightarrow 2\sqrt { x+1 } \sqrt { 3x+13 } =3x+12-5x-14=-2x-2$$
$$\Rightarrow 2\sqrt { x+1 } \sqrt { 4x+13 } =-2\left(x+1\right)$$
squaring both sides
$$\Rightarrow \left(x+1\right)\left(4x+13\right)=\left(x+1\right)^2=x^2+1+2x$$
     $$4x^2+4x+13+13x=x^2+2x+1$$
     $$3x^2+15x+12=0$$
     $$x^2+5x+4=0$$
     $$x^2+4x+x+4=0$$
     $$x\left(x+4\right)+\left(x+4\right)=0$$
     $$\left(x+4\right)+\left(x+4\right)=0$$
$$\Rightarrow x=-1,-4$$
Hence, the answer is $$-1,-4.$$


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