Unformatted text preview: ct that third term is a constant, means that this equation is reducible to
quadratic in form. We will solve this by first defining, u = x2
Now, this means that u 2 = ( x2 ) = x4
2 Therefore, we can write the equation in terms of u’s instead of x’s as follows, x 4  7 x 2 + 12 = 0 Þ u 2  7u + 12 = 0 The new equation (the one with the u’s) is a quadratic equation and we can solve that. In fact this
equation is factorable, so the solution is, u 2  7u + 12 = ( u  4 ) ( u  3) = 0 Þ u = 3, u = 4 So, we get the two solutions shown above. These aren’t the solutions that we’re looking for. We
want values of x, not values of u. That isn’t really a problem once we recall that we’ve defined u = x2
To get values of x for the solution all we need to do is plug in u into this equation and solve that
for x. Let’s do that. u = 3: 3 = x2 Þ x=± 3 u = 4: 4 = x2 Þ x = ± 4 = ±2 So, we have four solutions to the original equation, x = ±2 and x = ± 3 .
© 2007 Paul Dawkins 111 http://tutorial.math.lam...
View
Full
Document
 Spring '12
 MrVinh
 ........., Paul Dawkins

Click to edit the document details