5) The equation 2qx^2 + qx + 4p = 13x - 3 has the roots p and 1/q.?

Question

a) Find the values of p and q.

b) Hence, by using the values of p and q in a), form the quadratic equation which has roots p and -2q.

Answer 1

2qx^2 + qx + 4p = 13x - 3

2qx^2 + (q - 13) x + 4p + 3 = 0

1. substitute root p for x

then  2qp + ( q - 13 ) root p + 4p + 3 = 0

2pq + (q-13)(root p) + 4p + 3 = 0

2. substitute root 1/q for x

then 2 + ( q - 13) (1/q) + 4p + 3 = 0

2 + 1 - 13/q + 4p + 3 = 0

-13/q + 4p + 6 = 0

finally you have two equations

①2pq + (q-13)(root p) + 4p + 3 = 0

②-13/q + 4p + 6 = 0

② x q

=   -13 + 4pq + 6q = 0

=  q (4p + 6 - 13/q) = 0

it means   q = 0   or  4p + 6 - 13/q = 0

4p + 6 - 13/q = 0

4p = 13/q -6

p = 13/4q - 3/2

So  we know that q=0   or p = 13/4q - 3/2 from ②

and ①

①2pq + (q-13)(root p) + 4p + 3 = 0

I solved it up to that

and it took pretty much time

I tried to solve it

but I can't

I hate your math teacher

Answer 2

Part a)  Recast the quadratic into standard form:

2qx² + x(q-13) + (4p+3) = 0

This should somehow "match" the quadratic you get by identifying p and 1/q as roots:

(x-p) * (x - 1/q) = 0

Multiply the above by 2q so that you match the x² coefficients:

2 (x-p) (qx - 1) = 0 = 2qx² + x(-2 - 2pq) + 2p

Now equate coefficients (on x and the constant term):

(4p + 3) = 2p  =>

2p = -3  =>

p = -3/2

(q - 13) = -2 -2pq  =>

q(1 + 2p) = 11 =>

q(-2) = 11 =>

q = -11/2

Part b)  The quadratic is

(x-p)(x+2q) = 0  so we substitute p=-3/2, q=-11/2

(x+3/2)(x-11) = 0

x² - (19/2)x - 33/2 = 0  or you can multiply through by 2 if you don't like fractions:

2x² - 19x - 33 = 0

The moral of the exercise is that whenever someone tells you that you have a quadratic with roots r and s, you can immediately write down the equation for the quadratic as

(x-r)(x-s) = 0