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How does 2x squared minus x equal one half? - Answers

2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!



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How does 2x squared minus x equal one half? - Answers

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2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!



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https://math.answers.com/algebra/How_does_2x_squared_minus_x_equal_one_half

How does 2x squared minus x equal one half? - Answers

2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!

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      2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!
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