Find the square root of (√(50))+(√(48))


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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
	$We have ((√(50))+(√(48)))^2 =98+2(√(50.48)) =98+2(√(25.16.6))=98+40(√6) Hence (√((√(50))+(√(48))))=^4 (√(((√(50))+(√(48)))^2 )) =^4 (√(98+40(√6)))=^4 (√((a(√2)+b(√3))^4 )) (1) We need find a,b so that 98+40(√6)=(a(√2)+b(√3))^4 =4a^4 +8a^3 (√2).b(√3)+6.2a^2 .3b^2 +4.3b^3 (√3).a(√2) +9b^4 =(4a^4 +9b^4 +36a^2 b^2 +(8a^3 b+12ab^3 )(√6) ⇔ { ((4a^4 +36a^2 b^2 +9b^4 =98)),((8a^3 b+12ab^3 =40)) :} ⇔ { ((4a^4 +36a^2 b^2 +9b^4 =98)),((2a^3 b+3ab^3 =10(2))) :} ⇔10(4a^4 +36a^2 b^2 +9b^4 )=98(2a^3 b+3ab^3 ) Put a=kb substituting into above equality then divide two sides by k^4 we get 40k^4 +360k^2 +90=196k^3 +294k ⇔20k^4 −98k^3 +180k^2 −147k+45=0 This equation has one root k=1 ⇒a=b.Replace into (2)we get 5a^4 =10⇔a=^4 (√2).Replace into (1) we have (√((√(50))+(√(48))))=^4 (√(98+40(√6))) =^4 (√([^4 (√2)((√2)+(√3))]^4 )) =^4 (√2) ((√2)+(√3)) Thus, (√((√(50))+(√(48))))=^4 (√2) ((√2)+(√3))$
	$We have {(\sqrt{50} + \sqrt{48})}^{2} = 98 + 2 \sqrt{50.48}$ $= 98 + 2 \sqrt{25.16.6} = 98 + 40 \sqrt{6}$ $Hence \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{{(\sqrt{50} + \sqrt{48})}^{2}}$ $=^{4} \sqrt{98 + 40 \sqrt{6}} =^{4} \sqrt{{(a \sqrt{2} + b \sqrt{3})}^{4}} (1)$ $We need find a, b so that$ $98 + 40 \sqrt{6} = {(a \sqrt{2} + b \sqrt{3})}^{4}$ $= {4 a}^{4} + {8 a}^{3} \sqrt{2} . b \sqrt{3} + 6 . {2 a}^{2} . {3 b}^{2} + 4 . {3 b}^{3} \sqrt{3} . a \sqrt{2}$ $+ {9 b}^{4} = ({4 a}^{4} + {9 b}^{4} + {36 a}^{2} b^{2} + ({8 a}^{3} b + {12 ab}^{3}) \sqrt{6}$ $\Leftrightarrow {\begin{cases} {4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4} = 98 \\ {8 a}^{3} b + {12 ab}^{3} = 40 \end{cases}$ $\Leftrightarrow {\begin{cases} {4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4} = 98 \\ {2 a}^{3} b + {3 ab}^{3} = 10 (2) \end{cases}$ $\Leftrightarrow 10 ({4 a}^{4} + {36 a}^{2} b^{2} + {9 b}^{4}) = 98 ({2 a}^{3} b + {3 ab}^{3})$ $Put a = kb substituting into above$ $equality then divide two sides by k^{4}$ $we get {40 k}^{4} + {360 k}^{2} + 90 = {196 k}^{3} + 294 k$ $\Leftrightarrow {20 k}^{4} - {98 k}^{3} + {180 k}^{2} - 147 k + 45 = 0$ $This equation has one root k = 1$ $\Rightarrow a = b . Replace into (2) we get$ ${5 a}^{4} = 10 \Leftrightarrow a =^{4} \sqrt{2} . Replace into (1)$ $we have \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{98 + 40 \sqrt{6}}$ $=^{4} \sqrt{{[^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})]}^{4}} =^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})$ $Thus, \sqrt{\sqrt{50} + \sqrt{48}} =^{4} \sqrt{2} (\sqrt{2} + \sqrt{3})$
	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

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