Bunuel wrote:
How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?
A. 0
B. 1
C. 2
D. 3
E. 4
\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)
\(\sqrt{x^2 + 1} + \sqrt{x^2 + 1 + 1} = 2\)
let \({x^2 + 1}\) = y
\(\sqrt{y} + \sqrt{y + 1} = 2\)-------------------------------(1)
squaring both side
\({(\sqrt{y} + \sqrt{y + 1} )}^2\) = y + (y+1) + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 4
2y+1 + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 4
2y + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 3
2 (y + \((\sqrt{y} * \sqrt{y + 1} )\) ) = 3
2 \(\sqrt{y}\) ( \(\sqrt{y}+ \sqrt{y + 1}\) ) =3
2 \(\sqrt{y}\) (2) =3-------------------------------------------( from 1)
4 \(\sqrt{y}\) =3
\(\sqrt{y}\) =\(\frac{3}{4}\)
y = \({x^2 + 1}\)
\(\sqrt{{x^2 + 1}}\) =\(\frac{3}{4}\)
\({x^2 + 1}\) = \((\frac{3}{4})^2\)
\({x^2 + 1}\) = \(\frac{9}{16}\)
\({x^2 }\) =\(\frac{9}{16} -1\)
\({x^2 }\) = -ve value
so no Solution for X.
so ans is A