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

$if x + y = \sqrt{7} and x^{2} - y^{2} = \sqrt{35};$ $then find the value of 16 x y (x^{2} + y^{2})$
Commented by abdullah_ff last updated on 21/Nov/21

$is the answer 48 ?$
	Answered by Rasheed.Sindhi last updated on 21/Nov/21
	$x + y = (√7) ............(i) x^2 − y^2 = (√(35)) .........(ii) (ii)/(i):x−y=((x^2 −y^2 )/(x+y))=((√(35))/( (√7)))=(√5) ...(iii) (i)+(iii): 2x=(√7) +(√5) ⇒x=(((√7) +(√5))/2) (i)−(iii):2y=(√7) −(√5) ⇒y=(((√7) −(√5))/2) ▶16xy(x^2 +y^2 ) =16((((√7) +(√5))/2))((((√7) −(√5))/2)){((((√7) +(√5))/2))^2 +((((√7) −(√5))/2))^2 } 16(((7−5)/4))(((7+5+2(√7) (√5))/4)+((7+5−2(√7) (√5))/4)) 4(2)(((24)/4))=48$
	$x + y = \sqrt{7} . . . . . . . . . . . . (i)$ $x^{2} - y^{2} = \sqrt{35} . . . . . . . . . (i i)$ $(i i) / (i) : x - y = \frac{x^{2} - y^{2}}{x + y} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{5} . . . (i i i)$ $(i) + (i i i) : 2 x = \sqrt{7} + \sqrt{5} \Rightarrow x = \frac{\sqrt{7} + \sqrt{5}}{2}$ $(i) - (i i i) : 2 y = \sqrt{7} - \sqrt{5} \Rightarrow y = \frac{\sqrt{7} - \sqrt{5}}{2}$ $▸ 16 x y (x^{2} + y^{2})$ $= 16 (\frac{\sqrt{7} + \sqrt{5}}{2}) (\frac{\sqrt{7} - \sqrt{5}}{2}) {{(\frac{\sqrt{7} + \sqrt{5}}{2})}^{2} + {(\frac{\sqrt{7} - \sqrt{5}}{2})}^{2}}$ $16 (\frac{7 - 5}{4}) (\frac{7 + 5 + 2 \sqrt{7} \sqrt{5}}{4} + \frac{7 + 5 - 2 \sqrt{7} \sqrt{5}}{4})$ $4 (2) (\frac{24}{4}) = 48$
	Answered by Rasheed.Sindhi last updated on 21/Nov/21

	$\underset{(i)}{\underset{⏟}{x + y = \sqrt{7}}} \land \underset{(i i)}{\underset{⏟}{x^{2} - y^{2} = \sqrt{35}}}$ $(i i) / (i) : \frac{x^{2} - y^{2}}{x + y} = x - y = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{5}$ $\begin{matrix} \underset{\underset{2 (x^{2} + y^{2}) = 12 . . . . . . . A}{{(\sqrt{7})}^{2} + {(\sqrt{5})}^{2} = 2 (x^{2} + y^{2})}}{{(x + y)}^{2} + {(x - y)}^{2} = 2 (x^{2} + y^{2})} \end{matrix}$ $\begin{matrix} \underset{\underset{4 x y = 2 . . . . . . . B}{{(\sqrt{7})}^{2} - {(\sqrt{5})}^{2} = 4 x y}}{{(x + y)}^{2} - {(x - y)}^{2} = 4 x y} \end{matrix}$ $A \times B \times 2 :$ $16 x y (x^{2} + y^{2}) = 12 \times 2 \times 2 = 48$
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