Given info:- X = √(4-√7) and Y = √(4+√7)
Find the value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)²?
Explanation:-
Given that:
X = √(4-√7)
It can be written as
On multiplying and dividing by 2 in the squre root
⇛ X = √[2{(4-√7)}/2]
⇛ X =√[(8-2√7)/2]
⇛ X =√[{(7+1)-2√7{/2]
⇛ X =√[{(√7)²+(√1)²-2(√7)(√1)}/2]
It is in the form of a²-2ab+b²
Where , a = √7 and b = √1
⇛X = √[{(√7-√1)}²/2] [∵(a-b)² = a²-2ab+b²]
⇛ X = (√7-√1)/√2
⇛ X = (√7-1)/√2 — — — — Eqⁿ(1)
and
Y = √(4+√7)
It can be written as
On multiplying and dividing by 2 in the squre root
⇛ Y = √[{2(4+√7)}/2]
⇛ Y =√[(8+2√7)/2]
⇛ Y =√[{(7+1)+2√7}/2]
⇛ Y =√[{(√7)²+(√1)²+2(√7)(√1)}/2]
It is in the form of a²+2ab+b²
Where , a = √7 and b = √1
⇛Y = √[{(√7+√1)}²/2] [∵(a+b)² = a²+2ab+b²]
⇛ Y = (√7+√1)/√2
⇛ Y = (√7+1 )/√2 — — — — Eqⁿ(2)
On adding Eqⁿ (1) & Eqⁿ(2)
→ X+Y = (√7-1)/√2+(√7+1)/√2
→ X+Y = (√7-1+√7+1)/√2
→ X+Y = (2√7 )/√2
→ X+Y = (√2×√2×√7)/√2
→ X+Y = √2×√7
→ X+Y = √14 — — — — Eqⁿ(3)
On Subtracting Eqⁿ(2) from Eqⁿ(1)
→ X-Y =[(√7-1)/√2]-[(√7+1)/√2]
→ X-Y =[ (√7-1 )-(√7+1)]/√2
→ X-Y = (√7-1-√7-1)/√2
→ X-Y = (-1-1)/√2
→ X-Y = -2/√2
→ X-Y = -(√2×√2)/√2
→ X-Y = -√2 — — — — Eqⁿ(4)
On multiplying Eqⁿ(1)& Eqⁿ(2)
→ XY = [(√7-1)/√2][(√7+1)/√2]
→ XY = (√7-1)×(√7+1)/(√2×√2)
→ XY = [(√7)²-(1)²]/(2) [∵(a+b)(a-b) = a²-b²]
→ XY = (7-1)/2
→ XY = 6/2
→ XY = 3 — — — — — Eqⁿ(5)
Now,
The value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)²
It can be written as
⇛[(X-Y)²/3] + [ XY(X+Y) ]²+(X+Y)²
From Eqⁿ(3), Eqⁿ(4)& Eqⁿ(5)
⇛ [(-√2)²/3] + [ 3(√14)]²+(√14)²
⇛ (2/3) + (9×14)+(14)
⇛ (2/3)+(126)+(14)
⇛ (2/3)+140
⇛ [2+(3×140)]/3
⇛(2+420)/3
⇛ 422/3
Therefore, the value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)² is 422/3.
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