Quadratic equation

Question

The quadratic equation 2x²=6x+3 has roots p and q. Find the quadratic equation with the roots p²q dan pq².

Answer 1

To find roots p^2q and q^2p = 0,

(x-p^2q)(x-pq^2) = x^2 - (p^2q+q^2p)x + p^3q^3 = x^2 - pq(p+q)x + (pq)^3 = 0

Note that the original quadratic may be written (x-p)(x-q) = x^2 - (p+q)x + pq = 0

Thus, we see p+q and pq in the second equation, which means that they may be substituted into the top equation. It is the 2 in the original equation which must be divided out.

2x^2 - 6x - 3 = 0

x^2 - 3x - 3/2 = 0

Substituting into x^2 - (p+q)x + pq = 0,

p+q = 3

pq = -3/2

Then, substituting into the top equation, x^2 - pq(p+q)x + (pq)^3

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

x^2 + 9/2x - 27/8 = 0

To remove the fraction,

8x^2 + 36x - 27 = 0

Answer 2

2x^2 - 6x - 3 =0

this would be a simpler problem if it were 5 instead of 6 for the linear term's coefficient

maybe you have a typo? 5 and 6 are close on the keyboard

2x^2-5x-3 has integer factors

(2x+1)(x-3)=0

then x=3 or -1/2

p=3, q=--1/2

p^2q = -9/2

pq^2 = 3/4

(x+9/2)(x-3/4)

x^2 +15/4)x -27/8 = 0 or

8x^2 +30x -27 = 0 or

8x^2 +30x = 27 is a quadratic equation with roots p^2q and pq^2

2x^2 -6x -3 =0 is more complicated

x= 6/4 + or - (1/4)sqr(36+24) = 3/2 + or - (1/2)sqr15

p= 3/2 + sqr15/2

q =3/2- sqr15/2

p^2 = 9/4 + (3/2)sqr15 + 15/4

p^2q = (9/4+(3/2)sq15 +15/4)(3/2-sqr15/2)

pq^2 = (3/2 +sqr15/2)(9/4 -(3/2)sqr15 +15/4)

(x-p^2q)(x+pq^2) = 0 is an equation with roots p^2q and pq^2. "just" substitute the above values for p and q

instructors vary. some give major partial credit for working a problem that you slightly alter. Others give no credit. others give full credit & extra credit, when they realize they made a mistake in the problem and you found what the problem really was.

But there has to be easier ways to do this

Notice that 2x^2-5x-3 converts to 8x2-30x-27 by increasing coefficients and constant by factors 4,6 and 9

do the same with 2x^2-6x-3 to get:

8x^2-36x-27=0 as the quadratic with roots p^2q and pq^2