If x= 2-root 3
find x+1whole square - x-1whole square
pl help
thanks
kumaravel
karthik4297:
is your question (x+1)^2 - (x-1)^2
Answers
Answered by
1
Important formulas:-
(a + b)² + (a - b)² = 2a² + 2b²
(a + b)² - (a - b)² = 4ab
Above formulas should be remembered . Also should confirm above result by expanding (a + b)² and (a - b)² and simplifying. This will be an aid to memorise also as you will know how these results are obtained.
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Now to the question.
you have to find, (x + 1)² - (x - 1)²
If you compare with the above formulas , see that a = x and b = 1,
Hence from second formula, (x + 1)² - (x - 1)² = 4.x.1 = 4x = 4(2 - √3) = 8 - 4√3
(a + b)² + (a - b)² = 2a² + 2b²
(a + b)² - (a - b)² = 4ab
Above formulas should be remembered . Also should confirm above result by expanding (a + b)² and (a - b)² and simplifying. This will be an aid to memorise also as you will know how these results are obtained.
-----------------------------------------------------
Now to the question.
you have to find, (x + 1)² - (x - 1)²
If you compare with the above formulas , see that a = x and b = 1,
Hence from second formula, (x + 1)² - (x - 1)² = 4.x.1 = 4x = 4(2 - √3) = 8 - 4√3
Answered by
1
x=2-√3
so,x+1=2-√3+1
=3-√3
so,(x+1)²=(3-√3)²
By using the identity (a-b)²=a²-2ab+b²,we get
(3-√3)²=3²-2*3*√3+(√3)²
=9-6√3+3
=12-6√3
=6(2-√3) (FOUND SAME AS 6x)
ANS=6(2-√3)
so,x+1=2-√3+1
=3-√3
so,(x+1)²=(3-√3)²
By using the identity (a-b)²=a²-2ab+b²,we get
(3-√3)²=3²-2*3*√3+(√3)²
=9-6√3+3
=12-6√3
=6(2-√3) (FOUND SAME AS 6x)
ANS=6(2-√3)
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