# If y=sqrt(x^2+6x+8),show that one value of sqrt(1+iy)+sqrt(1-iy)=sqrt(2x+8)?

Dec 30, 2017

Kindly find a Proof in the Explanation.

#### Explanation:

$y = \sqrt{{x}^{2} + 6 x + 8}$.

$\therefore {y}^{2} = {x}^{2} + 6 x + 8$.

Adding $1$ to both sides, ${y}^{2} + 1 = {x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$.

Taking square root, $\sqrt{{y}^{2} + 1} = x + 3 , \mathmr{and} , x = \sqrt{{y}^{2} + 1} - 3$

$\text{Therefore, } 2 x + 8 ,$

$= 2 \left\{\sqrt{{y}^{2} + 1} - 3\right\} + 8$,

$= 2 \sqrt{{y}^{2} + 1} + 2$,

$= 2 + 2 \sqrt{1 - \left(- {y}^{2}\right)}$,

$= 2 + 2 \sqrt{1 - {\left(i y\right)}^{2}}$,

=2+2sqrt{(1+iy)(1-iy),

=(1+iy)+(1-iy)+2sqrt{(1+iy)(1-iy),

$= {\left(\sqrt{1 + i y}\right)}^{2} + {\left(\sqrt{1 - i y}\right)}^{2} + 2 \sqrt{\left(1 + i y\right) \left(1 - i y\right)} , i . e . ,$,

$2 x + 8 = {\left\{\sqrt{1 + i y} + \sqrt{1 - i y}\right\}}^{2}$.

Taking square root, we get,

$\sqrt{2 x + 8} = \sqrt{1 + i y} + \sqrt{1 - i y} ,$ as desired!

Q.E.D.

Enjoy Maths.!

Dec 30, 2017

see below

#### Explanation:

sqrt(1+iy)+sqrt(1−iy)=sqrt(2x+8)

1+iy+2sqrt(1+iy)sqrt(1−iy)+1-iy=2x+8

2+2sqrt(1+iy)sqrt(1−iy)=2x+8

sqrt((1+iy)*(1−iy))=(2x+6)/2

(sqrt(1−(iy)^2))^2=(x+3)^2

$1 + {y}^{2} = {x}^{2} + 6 x + 9$

${y}^{2} = {x}^{2} + 6 x + 8$

$y = \sqrt{{x}^{2} + 6 x + 8}$

Now we know it is true for any number from interval: $x \in \left(- \infty , - 4\right] \bigcup \left[- 2 , \infty\right)$