What would the solution to this equation be using the quadratic formula
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What would the solution to this equation be using the quadratic formula

[From: ] [author: ] [Date: 11-11-29] [Hit: ]
Therefore,The squareroot of 40 is 2*sqrt(10).Therefore, there would still be a denominator of 2.......
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.

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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

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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.

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2x^2-8x+3=o

Δ=64-24=40

x = (8+/-2sqrt10)/4
x=2+/-(sqrt10)/2
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