(x+5)square-(x+2)square=-2 solve
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3
(x+5)^2 -(x+2)^2 = -2
[ (a+b)^2 = a^2 + b^2 +2ab ]
x^2 + 25 + 10x - (x^2 +4 + 4x)=-2
x^2 + 25 + 10x - x^2 -4 -4x=-2
25 -4 + 10x -4x=-2
21 + 6x = -2
6x = -2 -21
6x = -23
x = -23/6
[ (a+b)^2 = a^2 + b^2 +2ab ]
x^2 + 25 + 10x - (x^2 +4 + 4x)=-2
x^2 + 25 + 10x - x^2 -4 -4x=-2
25 -4 + 10x -4x=-2
21 + 6x = -2
6x = -2 -21
6x = -23
x = -23/6
Sanjana5428:
ok,i will try to solve
thanks
once u check weather it is (x-2)square or (x+2)square
in ur book
x+2
if the question is correct then this will only be the answer
fine ill try again
other 2 people r watching, think they may answer this
in place of (x+2)square if we take (x-2)square then we will get 1*9/14
no its x+2 only
Answered by
2
(x+5)^2-(x+2)^2=-2
In LHS apply identity a^2-b^2=(a+b)(a-b).... Here a= x+5 and b= x+2..
=> (x+5+x+2)(x+5-x-2)= -2
=>( 2x+7) 3= -2
=> 6x+21= -2
=> 6x= -2-21
=> x= -23/6
Plz mark brainliest
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