Solving this system of quadratic equations: determine a, b, and c in terms of p's and q's.?

Question

The graph of the function f(x)=ax^2+bx+c contains the points (p-q,q), (p,-2q), and (p+q, q) with q not equal to 0. Determine a, b, and c in terms of p's and q's
thanks in advance!

Answer 1

Plugging the data into the equation f(x) = a*x^2 + bx + c gives us the following:



(i) (p-q, q): q = a*(p-q)^2 + b*(p-q) + c

(ii) (p, -2q): -2q = a*p^2 + bp + c

(iii) (p+q, q): q = a*(p+q)^2 + b*(p+q) + c



Equating (i) and (iii) gives us



a*(p-q)^2 + b*(p-q) + c = a*(p+q)^2 + b*(p+q) + c ---->



b*[(p-q) - (p+q)] = a*[(p+q)^2 - (p-q)^2] ---->



-2b*q = a*4pq ----> b = -2ap, since q is not equal to 0.



Plug this into equation (ii) to get -2q = a*p^2 - 2a*p^2 + c ----> c = a*p^2 - 2q.



Now plug b = -2ap and c = a*p^2 - 2q into equation (i) to find that



q = a*(p - q)^2 - 2ap*(p - q) + a*p^2 - 2q ---->



3q = a*[(p-q)^2 - 2p*(p-q) + p^2] = a*[p^2 - 2pq + q^2 - 2*p^2 + 2pq + p^2] ---->



3q = a*q^2 ----> a = 3/q, and so b = -6*p/q and c = (3*p^2 /q) - 2q.

Answer 2

Given that : f(x) = ax^2+bx+c, and points A (p-q,q), C(p,-2q), & D(p+q, q).



point A(p-q, q), f(x) = q = a(p-q)^2 +b(p-q) +c ........................................ [1]



point B(p-q, q), f(x) = -2q = ap^2 +bp +c .......................................... [2]



point C(p-q, q), f(x) = q = a(p+q)^2 +b(p+q) +c ...................................... [3]



Subtract [1] from [3], a[(p+q)^2 -(p-q)^2] +b[p+q-p+q] = 0,

OR,

4pqa + 2qb = 0, ==> b = -2ap ........................................................ [4]



Adding [1] & [3], 2q = a[(p+q)^2+(p-q)^2] +b[p+q + p-q] +2c

OR,

a(p^2+q^2) + bp + c = 2q ............................................................... [5]



Adding [2] & [5], a(p^2+p^2+q^2) +b(p+p) +2c = 0,

OR,

a(2p^2+q^2) + 2bp +2c = 0 ....................................................... [6]



From [4] & [6], a(2p^2+q^2) +2p(-2ap) +2c = 0,

OR,

a[2p^2+q^2 -4p^2] +2c = 0, ===> c = -a[q^2-2p^2]/2 ............. [7]



From [2], [4] & [7], ap^2 +p(-2ap) -a(q^2-2p^2)/2 = -2q,

OR,

a[p^2-2p^2-q^2/2+p^2] = -2q, ==> -aq^2/2 = -2q,

Thus,

a = 4 / q >=================================< ANSWER,



b = -2p*(4/q) = -8p/q >==========================< ANSWER

AND

c = (-4/q)[q^2-2p^2]/2 = 2(2p^2-q^2)/q >=============< ANSWER

Answer 3

Substitute the given points to x and y.





(1) q = a(p – q)² + b(p - q) + c



(2) –2q = a(p)² + b(p) + c



(3) q = a(p + q) ² + b(p + q) + c





Solve for a, b, and c. Subtracting (2) from (1) leads to





q + 2q = a(p – q)² – ap² + b(p - q) – bp



 3q = a[(p – q)² – p] + b[(p – q) – p]



 3q = a[(p² – 2pq + q²) – p²] + b(p – q – p)



 3q = a(p² – 2pq + q² – p²) + b(–q)



 3q = a(–2pq + q²) – bq



 3q = –2apq + aq² – bq (4)





Subtracting (2) from (3) leads to





q + 2q = a(p + q)² – ap² + b(p + q) – bp



 3q = a[(p + q)² – p²] + b[(p + q) – p]



 3q = a[(p² + 2pq + q²) – p²] + b(p + q – p)



 3q = a(p² + 2pq + q² - p²) + b(q)



 3q = a(2pq + q²) + bq



 3q = 2apq + aq² + bq (5)





Adding (4) and (5) leads to





6q = 2aq²



 a = 6q/2q²



 a = 3/q





Substitute 3/q to a in (5) and solve for b.





3q = 2(3/q)pq + (3/q)q² + bq



 3q = 6p + 3q + bq



 3p = 9p + bq



 bq = – 6p



 b = – 6p/q





Substitute 3/q to a and – 6p/q to b in (2) and solve for c.





–2q = (3/q)(p)² + (– 6p/q)(p) + c



 –2q = 3p²/q – 6p²/q + c



 –2q = –3p²/q + c



 c = 3p²/q – 2q



 c = (3p² – 2q²)/q





Therefore, a = 3/q, b = – 6p/q, and c = (3p² – 2q²)/q.