Question #253d6

1 Answer
Oct 16, 2015

Answer:

Here's what I got.

Explanation:

You know that two quadratic equations

#3x^2 + px + 1 = 0" " " "color(purple)((1))#

and

#2x^2 + qx + 1 = 0" " " "color(purple)((2))#

have a common root and that the following relationship exists between #p# and #q#

#2p^2 + 3q^2 -5pq + 1 = 0#

Now let's assume that #color(blue)(n)# is the common root. Since that would imply that #color(blue)(n)# is a root for both equations, you can say that

#3 * color(blue)(n)^2 + p * color(blue)(n) + 1 = 0#

and

#2 * color(blue)(n)^2 + q * color(blue)(n) + 1 = 0#

This is equivalent to saying that

#3n^2 + pn + color(red)(cancel(color(black)(1))) = 2n^2 + qn + color(red)(cancel(color(black)(1)))#

Rearrange to get the value of #n# in terms of #p# and #q#

#3n^2 - 2n^2 + pn - qn = 0#

#n^2 + n(p-q) = 0#

#n * (n + p-q) = 0 implies {(n=0), (n + p - q = 0 <=> n = q - p) :}#

SInce #n=0# does not satify the original equations, you get that #n = q - p#.

Check to see if the given relationship is true, first for equation #color(purple)((1))#

#3 * (q-p)^2 + p * (q-p) + 1 = 0#

#3q^2 - 6pq + 3p^2 + pq - p^2 + 1 = 0#

This is equal to

#2p^2 + 3q^2 - 5pq + 1 = 0color(white)(x)color(green)(sqrt())#

and the nfor equation #color(purple)((2))#

#2(q-p)^2 + q(q-p) + 1 = 0#

#2q^2 - 4pq + 2p^2 + q^2 - qp + 1 = 0#

This will once again give

#2p^2 + 3q^2 - 5pq + 1 = 0color(white)(x)color(green)(sqrt())#