Question

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

Answer 1

Kindly find a Proof in the Explanation.

Explanation:

`y=sqrt(x^2+6x+8)`.

`:. y^2=x^2+6x+8`.

Adding `1` to both sides, `y^2+1=x^2+6x+9=(x+3)^2`.

Taking square root, `sqrt(y^2+1)=x+3, or, x=sqrt(y^2+1)-3`

`"Therefore, "2x+8,`

`=2{sqrt(y^2+1)-3}+8`,

`=2sqrt(y^2+1)+2`,

`=2+2sqrt(1-(-y^2))`,

`=2+2sqrt(1-(iy)^2)`,

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

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

`=(sqrt(1+iy))^2+(sqrt(1-iy))^2+2sqrt{(1+iy)(1-iy)}, i.e.,`,

`2x+8={sqrt(1+iy)+sqrt(1-iy)}^2`.

Taking square root, we get,

`sqrt(2x+8)=sqrt(1+iy)+sqrt(1-iy),` as desired!

Q.E.D.

Enjoy Maths.!

Answer 2

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+6x+9`

`y^2=x^2+6x+8`

`y=sqrt(x^2+6x+8)`

Now we know it is true for any number from interval: `x in (-oo,-4]uuu[-2,oo)`