Solveeit Logo
Login

Question

Question: ‌‌If we are given an expression as \[y=\sqrt{{{x}^{2}}+6x+8}\],‌ ‌show‌ ‌that‌ ‌one‌ ‌value‌ ‌of‌ ‌ ...

‌‌If we are given an expression as y=x2+6x+8y=\sqrt{{{x}^{2}}+6x+8},‌ ‌show‌ ‌that‌ ‌one‌ ‌value‌ ‌of‌ ‌ 1+iy+1−iy=2x+8\sqrt{1+iy}+\sqrt{1-iy}=\sqrt{2x+8}?‌

Explanation

Solution

From the given question we have to show that one value of ⇒[1+iy+1−iy]=2x+8\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=x2+6x+8y=\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
⇒y2=x2+6x+8\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,
⇒x2+6x+8−y2=0\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
⇒x=−6±36−4(8−y2)2\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
⇒x=−6±36−32+4y22\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
⇒2x+6=±36−32+4y21\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
⇒2x+6=±21+y2\Rightarrow 2x+6=\pm 2\sqrt{1+{{y}^{2}}}
From adding the 2 on both the sides the equation will be as follows
⇒2x+8=2+21+y2\Rightarrow 2x+8=2+2\sqrt{1+{{y}^{2}}}
⇒2x+8=1+iy+1−iy+2(1+iy)(1−iy)\Rightarrow 2x+8=1+iy+1-iy+2\sqrt{\left( 1+iy \right)\left( 1-iy \right)}
⇒2x+8=[1+iy+1−iy]2\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. ⇒[1+iy+1−iy]=2x+8\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 i2=−1{{i}^{2}}=-1rather that this if we use i2=1{{i}^{2}}=1then we can get1+y2=(1+iy)(1−iy)\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.