Look at the equation and if you can quickly determine

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Unformatted text preview: f eliminating the square root from the denominator so let’s also do that, x = -1 ± 6 5 5 = -1 ± 5 ( 6 )( 5) 5 = -1 ± 30 5 If you do clear the fractions out and run through the quadratic formula then you should get exactly the same result. For the practice you really should try that. [Return to Problems] (b) In this case do not get excited about the decimals. The quadratic formula works in exactly the same manner. Here are the values and the quadratic formula work for this problem. a = 0.04 x= b = -0.23 - ( -0.23) ± c = 0.09 ( -0.23) - 4 ( 0.04 )( 0.09 ) 2 ( 0.04 ) 2 0.23 ± 0.0529 - 0.0144 0.08 0.23 ± 0.0385 = 0.08 = © 2007 Paul Dawkins 102 http://tutorial.math.lamar.edu/terms.aspx College Algebra Now, to this will be the one difference between these problems and those with integer or fractional coefficients. When we have decimal coefficients we usually go ahead and figure the two individual numbers. So, let’s do that, x= 0.23 ± 0.0385 0.23 ± 0.19621 = 0.08 0.08 0.23 + 0.19621 0.08 = 5.327625 x= and and 0.23 - 0.19621 0.08 = 0.422375 x= N...
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