Solve p and q: X^2 - 6X + 10 = (X - p)^2 + q?

Question

X is an unknown. Find p and q.

THanx a lot i have been struggling with my math work today.

Answer 1

This is called 'completing the square'. So, to start you divide -6 by two to get -3. This goes in the brackets.

(X - 3 ) ^ 2 + q = X^2 - 6X + 10

Now you expand the brackets.

X^2 - 6X + 9 + q = X^2 - 6X + 10

To make both sides equal, you need to add 1. So, q would equal 1.

So, p = 3, and q = 1.

NOTE: p is not -3 because -(-3) is the equivalent of 3. This would make the equation wrong. The expansion would come out as X^2 + 6X + 10, not X^2 - 6X + 10

Answer 2

X^2 -6X +10= (X - 3)^2 + 1= (X-P)^2 +Q

Hence P= 3 and Q=1

Answer 3

the method of solution is called "completing the square"
x^2 - 6x + 10 = 0
move the constant to the rhs
x^2 - 6x = -10
find half the coefficent of x, square it, add to both sides
x^2 - 6x + 9 = -10 +9
the lhs can now be written as a perfect square
(x-3)^2 = -1
hence p=3 and q= (-1)

Answer 4

X^2 - 6X + 10 = (X - p)^2 + q

X^2 - 6X + 9 + 1 = (X - p)^2 + q

(X - 3)^2 + 1 = (X - p)^2 + q

Equating we get,

p = 3 and q =1

Answer 5

X^2 - 6X + 10 = (X - p)^2 + q
= X^2 -2pX +p^2 +q
[-6]X + 10 = [-2p]X + (p^2+q)
-2p = -6
p = 3
10 = (p^2+q)
10 = (3)^2+q
1 = q
p=3 and q =1

Answer 6

= (x² - 6x + 9) - 9 + 10
= (x - 3)² + 1
p = 3 , q = 1

Answer 7

(X-p)^2+q=X^2+2pX+P^2+q=X^2-6X+10 ==> 2p=-6,p^2+q=10 ==>
p=-3,9+q=10 ==> p=-3,q=1

Answer 8

Do you mean p = x + (q/y)+ q? If this is so, then p - x = q(1+y)/y, so q a q = y(p - x)/(y + 1)

Answer 9

x^2-6X+10=X^2+p^2-2px+q
compare the coefficient,
u get,
2p=6.......1
p^2+q=10........2
solve, by equating both 1 and 2
u'll get p and q