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‌‌If we are given an expression as \[y=\sqrt{{{x}^{2}}+6x+8}\],‌ ‌show‌ ‌that‌ ‌one‌ ‌value‌ ‌of‌ ‌ \[\sqrt{1+iy}+\sqrt{1-iy}=\sqrt{2x+8}\]?‌

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Last updated date: 20th Jul 2024
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Answer
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Hint: From the given question we have to show that one value of \[\Rightarrow \left[ \sqrt{1+iy}+\sqrt{1-iy} \right]=\sqrt{2x+8}\].Here in this problem we have manipulate and get the proof by using the basic mathematics and having a general knowledge in regarding the square roots and the complex numbers.
\[y=\sqrt{{{x}^{2}}+6x+8}\]

Complete step-by-step solution:
By squaring the given question on both the right hand and left hand side of the equation the equation becomes
\[\Rightarrow {{y}^{2}}={{x}^{2}}+6x+8\]
By bringing the whole terms in left hand side to right side of the equation that is the whole terms to one side we get,
\[\Rightarrow {{x}^{2}}+6x+8-{{y}^{2}}=0\]
Let in the equation we consider the y will be an constant and so after assuming it as an constant it will become a quadratic equation so by basic formula of quadratic equation the roots can be expressed as follows
\[\Rightarrow x=\dfrac{-6\pm \sqrt{36-4\left( 8-{{y}^{2}} \right)}}{2}\]
By the general arithmetic simplifications as addition and multiplication the equation becomes as follows
\[\Rightarrow x=\dfrac{-6\pm \sqrt{36-32+4{{y}^{2}}}}{2}\]
By general simplification sending the 6 to other side after multiplying the x with two the equation becomes as follows
\[\Rightarrow 2x+6=\dfrac{\pm \sqrt{36-32+4{{y}^{2}}}}{1}\]
From solving the terms or expression inside the square root the equation becomes as follows
\[\Rightarrow 2x+6=\pm 2\sqrt{1+{{y}^{2}}}\]
From adding the 2 on both the sides the equation will be as follows
\[\Rightarrow 2x+8=2+2\sqrt{1+{{y}^{2}}}\]
\[\Rightarrow 2x+8=1+iy+1-iy+2\sqrt{\left( 1+iy \right)\left( 1-iy \right)}\]
\[\Rightarrow 2x+8={{\left[ \sqrt{1+iy}+\sqrt{1-iy} \right]}^{2}}\]
Appling the square root on both sides the equation becomes as follows and as required in the question. \[\Rightarrow \left[ \sqrt{1+iy}+\sqrt{1-iy} \right]=\sqrt{2x+8}\].

Note: We must be very careful in doing calculations and must be having knowledge regarding the finding the roots of a quadratic equations and must not do mistakes or else it would end up with total misleading solution example here we use \[{{i}^{2}}=-1\]rather that this if we use \[{{i}^{2}}=1\]then we can get\[\sqrt{1+{{y}^{2}}}=\sqrt{\left( 1+iy \right)\left( 1-iy \right)}\] in our solution. So we must have good touch in complex numbers.