The equation is 2x^2-8x+3=o. My precalculus book tells me it would be 2 +/- the square root of 10/2, but that makes no sense considering the step before you get your answer you have 8 +/- the square root of 40/ all over 4. So wouldn't the answer be 2 +/- the square root of 10? How would there still be a denominator of 2 when the denominator divides equally into 40?
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I got an answer of 2 +/- 2square root 10.
The 40 was under the square root so it can't be divided by the 2 until after square roots are factored out.
The 40 was under the square root so it can't be divided by the 2 until after square roots are factored out.
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Quadratic formula: [-b +/- √(b^2 - 4ac)] / 2a
2x^2 + (-8)x + 3 = 0 ==> ax^2 + bx +c = 0
Therefore, using the quadratic formula: [-(-8) +/- √((-8)^2 - 4(2)(3))] / (2x2)
= [ 8 +/- √(64-24)] / 4
= 8/4 +/- [√40]/4
= 2 +/- [√4√10]/4
= 2 +/- [2√10]/4 = 2 +/- √10/2
2x^2 + (-8)x + 3 = 0 ==> ax^2 + bx +c = 0
Therefore, using the quadratic formula: [-(-8) +/- √((-8)^2 - 4(2)(3))] / (2x2)
= [ 8 +/- √(64-24)] / 4
= 8/4 +/- [√40]/4
= 2 +/- [√4√10]/4
= 2 +/- [2√10]/4 = 2 +/- √10/2
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You have the following:
8+/-sqrt(40)/4
The squareroot of 40 is 2*sqrt(10). Therefore, there would still be a denominator of 2.
8+/-sqrt(40)/4
The squareroot of 40 is 2*sqrt(10). Therefore, there would still be a denominator of 2.
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2x^2-8x+3=o
Δ=64-24=40
x = (8+/-2sqrt10)/4
x=2+/-(sqrt10)/2
Δ=64-24=40
x = (8+/-2sqrt10)/4
x=2+/-(sqrt10)/2