Find-the-square-root-of-50-48-

Question Number 113913 by Aina Samuel Temidayo last updated on 16/Sep/20

Find the square root of \sqrt{50} + \sqrt{48}

Answered by 1549442205PVT last updated on 16/Sep/20

We have {(\sqrt{50} + \sqrt{48})}^{2} = 98 + 2 \sqrt{50.48}

= 98 + 2 \sqrt{25.16.6} = 98 + 40 \sqrt{6}

Hence \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{{(\sqrt{50} + \sqrt{48})}^{2}}

=^{4} \sqrt{98 + 40 \sqrt{6}} =^{4} \sqrt{{(a \sqrt{2} + b \sqrt{3})}^{4}} (1)

We need find a, b so that

98 + 40 \sqrt{6} = {(a \sqrt{2} + b \sqrt{3})}^{4}

= {4 a}^{4} + {8 a}^{3} \sqrt{2} . b \sqrt{3} + 6 . {2 a}^{2} . {3 b}^{2} + 4 . {3 b}^{3} \sqrt{3} . a \sqrt{2}

+ {9 b}^{4} = ({4 a}^{4} + {9 b}^{4} + {36 a}^{2} b^{2} + ({8 a}^{3} b + {12 ab}^{3}) \sqrt{6}

\Leftrightarrow {\begin{cases} {4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4} = 98 \\ {8 a}^{3} b + {12 ab}^{3} = 40 \end{cases}

\Leftrightarrow {\begin{cases} {4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4} = 98 \\ {2 a}^{3} b + {3 ab}^{3} = 10 (2) \end{cases}

\Leftrightarrow 10 ({4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4}) = 98 ({2 a}^{3} b + {3 ab}^{3})

Put a = kb substituting into above

equality then divide two sides by k^{4}

we get {40 k}^{4} + {360 k}^{2} + 90 = {196 k}^{3} + 294 k

\Leftrightarrow {20 k}^{4} - {98 k}^{3} + {180 k}^{2} - 147 k + 45 = 0

This equation has one root k = 1

\Rightarrow a = b . Replace into (2) we get

{5 a}^{4} = 10 \Leftrightarrow a =^{4} \sqrt{2} . Replace into (1)

we have \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{98 + 40 \sqrt{6}}

=^{4} \sqrt{{[^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})]}^{4}} =^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})

Thus, \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})

Commented by Aina Samuel Temidayo last updated on 16/Sep/20

Thanks .

Answered by mr W last updated on 16/Sep/20

\sqrt{50} + \sqrt{48}

= \sqrt{2} (5 + 2 \sqrt{6})

= \sqrt{2} [{(\sqrt{2})}^{2} + 2 \sqrt{2} \sqrt{3} + {(\sqrt{3})}^{2}]

= \sqrt{2} {[\sqrt{2} + \sqrt{3}]}^{2}

\Rightarrow \sqrt{\sqrt{50} + \sqrt{48}} = \sqrt[4]{2} (\sqrt{2} + \sqrt{3})

Answered by Dwaipayan Shikari last updated on 16/Sep/20

G e n e r a l l y \sqrt{x} + \sqrt{y} = \sqrt{a + \sqrt{b}}

x + y + 2 \sqrt{x y} = a + \sqrt{b}

(x + y) = a

4 x y = b

(x - y) = \sqrt{a^{2} - b}

\sqrt{x} + \sqrt{y} = \frac{\sqrt{a + \sqrt{a^{2} - b}}}{\sqrt{2}} + \sqrt{\frac{a - \sqrt{a^{2} - b}}{2}}

= \sqrt{\frac{\sqrt{50} + \sqrt{2}}{2}} + \sqrt{\frac{\sqrt{50} - \sqrt{2}}{2}} a = \sqrt{50} b = 48

Commented by Dwaipayan Shikari last updated on 16/Sep/20

\sqrt{\sqrt{49} + \sqrt{48}} = \sqrt{\frac{\sqrt{49} + 1}{2}} + \sqrt{\frac{\sqrt{49} - 1}{2}} = 2 + \sqrt{3} = \sqrt{\frac{8}{2}} + \sqrt{\frac{6}{2}}

\sqrt{\sqrt{64} + \sqrt{63}} = \sqrt{\frac{8 + 1}{2}} + \sqrt{\frac{8 - 1}{2}} = \sqrt{\frac{9}{2}} + \sqrt{\frac{7}{2}}

\sqrt{\sqrt{100} + \sqrt{99}} = \sqrt{\frac{11}{2}} + \sqrt{\frac{9}{2}}

\sqrt{\sqrt{81} + \sqrt{80}} = \sqrt{\frac{10}{2}} + \sqrt{\frac{9}{2}}

H a v e y o u n o t i c e d a n y p a t t e r n ?

\sqrt{1} = \sqrt{\frac{2}{2}} + \sqrt{\frac{0}{2}}

\sqrt{\sqrt{4} + \sqrt{3}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}

\sqrt{\sqrt{9} + \sqrt{8}} = \sqrt{\frac{4}{2}} + \sqrt{\frac{2}{2}}

\sqrt{\sqrt{16} + \sqrt{15}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{3}{2}}

\sqrt{\sqrt{25} + \sqrt{24}} = \sqrt{\frac{6}{2}} + \sqrt{\frac{4}{2}}

\sqrt{\sqrt{36} + \sqrt{35}} = \sqrt{\frac{7}{2}} + \sqrt{\frac{5}{2}}

\sqrt{\sqrt{n^{2}} + \sqrt{n^{2} - 1}} = \sqrt{\frac{n + 1}{2}} + \sqrt{\frac{n - 1}{2}}

\sqrt{n + \sqrt{n^{2} - 1}} = \sqrt{\frac{n + 1}{2}} + \sqrt{\frac{n - 1}{2}}

Commented by Rasheed.Sindhi last updated on 16/Sep/20

A c t u a l l e a r n i n g f r o m y o u r

g e n e r a l m e t h o d! T h a n k y o u .