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if-x-y-7-and-x-2-y-2-35-then-find-the-value-of-16xy-x-2-y-2-

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Question Number 159777 by abdullah_ff last updated on 21/Nov/21
if x + y = (√7) and x^2 − y^2 = (√(35)) ; then find the value of 16xy(x^2 + y^2 )
$$\mathrm{if}\:{x}\:+\:{y}\:=\:\sqrt{\mathrm{7}}\:\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\sqrt{\mathrm{35}}\:; \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{16}{xy}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \right) \\ $$
Commented by abdullah_ff last updated on 21/Nov/21
is the answer 48?
$$\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{48}? \\ $$
Answered by Rasheed.Sindhi last updated on 21/Nov/21
x + y = (√7) ............(i) x^2 − y^2 = (√(35)) .........(ii) (ii)/(i):x−y=((x^2 −y^2 )/(x+y))=((√(35))/( (√7)))=(√5) ...(iii) (i)+(iii): 2x=(√7) +(√5) ⇒x=(((√7) +(√5))/2) (i)−(iii):2y=(√7) −(√5) ⇒y=(((√7) −(√5))/2) ▶16xy(x^2 +y^2 ) =16((((√7) +(√5))/2))((((√7) −(√5))/2)){((((√7) +(√5))/2))^2 +((((√7) −(√5))/2))^2 } 16(((7−5)/4))(((7+5+2(√7) (√5))/4)+((7+5−2(√7) (√5))/4)) 4(2)(((24)/4))=48
$${x}\:+\:{y}\:=\:\sqrt{\mathrm{7}}\:…………\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\sqrt{\mathrm{35}}\:………\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right):{x}−{y}=\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}+{y}}=\frac{\sqrt{\mathrm{35}}}{\:\sqrt{\mathrm{7}}}=\sqrt{\mathrm{5}}\:…\left({iii}\right) \\ $$$$\left({i}\right)+\left({iii}\right):\:\mathrm{2}{x}=\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{5}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left({i}\right)−\left({iii}\right):\mathrm{2}{y}=\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{5}}\:\Rightarrow{y}=\frac{\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\blacktriangleright\mathrm{16}{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{16}\left(\frac{\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left\{\left(\frac{\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \right\} \\ $$$$\mathrm{16}\left(\frac{\mathrm{7}−\mathrm{5}}{\mathrm{4}}\right)\left(\frac{\mathrm{7}+\mathrm{5}+\mathrm{2}\sqrt{\mathrm{7}}\:\sqrt{\mathrm{5}}}{\mathrm{4}}+\frac{\mathrm{7}+\mathrm{5}−\mathrm{2}\sqrt{\mathrm{7}}\:\sqrt{\mathrm{5}}}{\mathrm{4}}\right) \\ $$$$\mathrm{4}\left(\mathrm{2}\right)\left(\frac{\mathrm{24}}{\mathrm{4}}\right)=\mathrm{48} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Nov/21
x + y = (√7) _((i)) ∧ x^2 − y^2 = (√(35)) _((ii)) (ii)/(i):((x^2 −y^2 )/(x+y))=x−y=((√(35))/( (√7)))=(√5) determinant ((((x+y)^2 +(x−y)^2 =2(x^2 +y^2 )_(((√7))^2 +((√5))^2 =2(x^2 +y^2 )_(2(x^2 +y^2 )=12.......A) ) ))) determinant ((( (x+y)^2 −(x−y)^2 =4xy _(((√7))^2 −((√5))^2 =4xy_(4xy=2.......B) ) ))) A×B×2: 16xy(x^2 + y^2 )=12×2×2=48
$$\underset{\left({i}\right)} {\underbrace{\:{x}\:+\:{y}\:=\:\sqrt{\mathrm{7}}\:}}\:\:\:\wedge\:\:\:\:\underset{\left({ii}\right)} {\underbrace{{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\sqrt{\mathrm{35}}\:}} \\ $$$$\left({ii}\right)/\left({i}\right):\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}+{y}}={x}−{y}=\frac{\sqrt{\mathrm{35}}}{\:\sqrt{\mathrm{7}}}=\sqrt{\mathrm{5}} \\ $$$$\begin{array}{|c|}{\underset{\underset{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{12}…….{A}} {\left(\sqrt{\mathrm{7}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}} {\left({x}+{y}\right)^{\mathrm{2}} +\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}}\\\hline\end{array}\: \\ $$$$\begin{array}{|c|}{\:\:\:\:\:\:\underset{\underset{\mathrm{4}{xy}=\mathrm{2}…….{B}} {\left(\sqrt{\mathrm{7}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{4}{xy}}} {\left({x}+{y}\right)^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4}{xy}\:\:}\:\:\:\:}\\\hline\end{array} \\ $$$${A}×{B}×\mathrm{2}: \\ $$$$\:\mathrm{16}{xy}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \right)=\mathrm{12}×\mathrm{2}×\mathrm{2}=\mathrm{48} \\ $$

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