Find the all real roots of the equation (2x^2-3)^2-x^2=0
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Is the answer correct.?
yes
you just need to write x = + - ( 3 / 2, - 1 ) ,your method is correct.
OK Thanks
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( 2x^2 - 3 )^2 - x^2 = 0
4 x^4 + 9 - 12 x^2 - x^2 = 0
4 x^4 - 13 x^2 + 9 = 0
4 x^4 - 9 x^2 - 4 x^2 + 9 = 0
x^2 ( 4 x^2 - 9 ) - 1 ( 4 x^2 - 9 ) = 0
( x^2 - 1 ) ( 4 x^2 - 9 ) = 0
{ ( x^2 - 1^2 ) } { (2x)^2 - ( 3 )^2 } = 0
( x - 1 ) ( x + 1 ) ( 2x - 3 ) ( 2x + 3 ) = 0
x - 1 = 0, x + 1 = 0, 2x - 3 = 0, 2x + 3 = 0
x = 1 , x = -1 , x = 3 / 2 , x = - 3 / 2
hence, x = 1 , - 1 , 3 / 2 , - 3 /2 are all real roots of given polynomial.
【 hope it helps 】
4 x^4 + 9 - 12 x^2 - x^2 = 0
4 x^4 - 13 x^2 + 9 = 0
4 x^4 - 9 x^2 - 4 x^2 + 9 = 0
x^2 ( 4 x^2 - 9 ) - 1 ( 4 x^2 - 9 ) = 0
( x^2 - 1 ) ( 4 x^2 - 9 ) = 0
{ ( x^2 - 1^2 ) } { (2x)^2 - ( 3 )^2 } = 0
( x - 1 ) ( x + 1 ) ( 2x - 3 ) ( 2x + 3 ) = 0
x - 1 = 0, x + 1 = 0, 2x - 3 = 0, 2x + 3 = 0
x = 1 , x = -1 , x = 3 / 2 , x = - 3 / 2
hence, x = 1 , - 1 , 3 / 2 , - 3 /2 are all real roots of given polynomial.
【 hope it helps 】
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