Maths Polynomials-3

Question

Given that p² +6p=q² +6q and that p does not equal q, find the values of:

 

    (a) p+q

 

    (b) 4(p+q)²

Answer 1

Just a visual alternative

 

Consider that the graph of p² + 6p is a parabola opening up with a vertex (-3, -9) 

 

As you move away from the vertex to the left and right, the y-values (or the value of p² + 6p) are symmetrical (equal)

 

If you move one unit to the left of the vertex, you get p = -4

 

Let q = the value you get when you move to the right

 

If you move one unit to the right of the vertex, you get q = -2

 

The value of p² +6p (or q² + 6q) will be the same at these two points

 

To check:  if p = -4, then p² + 6p = -8

 

If q= -2, then q² + 6q = -8

 

(a) Looking at the first question, add p and q

 

-4 + -2 = -6

 

If you moved 10 units to the left and right for example, you would have p = -13 and q =7 

 

-13 + 7 = -6

 

These numbers will always sum to -6 because their average must be -3

 

 as you move equal distances to the left and right to select values for p and q, p + q will always equal = -6

 

(b) if p + q = -6, then 4(p + q)² =

                           

                 = 4(-6)²

                 

                 = 4(36)

 

                 = 144

 

 

 

Answer 2

p^2+6p=q^2+6q

p^2-q^2=6q-6p

(p-q)(p+q)=6(q-p)

p+q=6(q-p)/(p-q)

multiply both factors in the numerator by -1

p+q=-6(p-q)/(p-q)

p+q=-6

 

since p+q=-6,...

4(p+q)^2=4(-6)^2=4(36)=144

Answer 3

p² + 6p = q² + 6q

 

p² - q² + 6p - 6q = 0

 

(p+q)(p-q) + 6(p-q) = 0

 

(p-q)(p+q+6) = 0

 

So, p-q = 0 or p+q+6 = 0

 

Since p≠q, p-q≠0.  Therefore, p+q = -6.