Question

if x^2+bx=(x+p)^2+q, express p and q in terms of b????

Answer 1

The general idea is, if you have
\[ x^2 +bx = x^2+2x+c \]
you can match coefficients: the b in front of the x on the left side matches the 2 in front of the x on the right side
c on the right must match (the unwritten) zero on the left. so you can say
b=2, c=0

to do your problem, expand the right side, and match coefficients.

Answer 2

how does the bx equate to a 2x?

Answer 3

I posted an example. It is not your problem. But if we told
\[ x^2 +bx = x^2+2x+c \]
then we can match coefficients.

Answer 4

Have you multiplied out (x+p)^2 out ?

Answer 5

x^2+2px+p^2.

Answer 6

so what is your problem now?
x^2+bx= x^2 +2px +p^2+q

can you match coefficients?

Answer 7

my problem is still the same.

Answer 8

Maybe this idea does not make sense, but what you do for this problem is match the coefficient of the x term on the left with the coefficient of the x term on the right.

Answer 9

what form is the first equation written in?

Answer 10

x^2+bx= x^2 +2px +p^2+q

Answer 11

no the stuff on the LHS.

Answer 12

I am not following your question.
But for this problem:
if x^2+bx=(x+p)^2+q, express p and q in terms of b????
we expand the (x+p)^2 so that we get x^2 and x's on the right side:

x^2+bx= x^2 +2px +p^2+q

now we match coefficients of the x term and the constant term

Answer 13

so the x^2s cancel out

Answer 14

We are not "solving" in the normal sense. We are doing something slightly easier:
the numbers in front of the x^2 term, the x term, and the x^0 (constant term) must match to make the left equal to the right side.

Answer 15

so 1*x^2 on both sides is ok. but to make the left side equal the right, the b must match the 2p.

Answer 16

ok sorry I do not get this

Answer 17

thanks for trying to help me.

Answer 18

Think about this example

3x= bx
for all x. what is b? Not 1: 3x≠1x, (this is only true for x=0, but we want it true for any x). Think about it, and you see b=3 makes both sides equal for any x.

Answer 19

ok then

Answer 20

The is the first idea. The second idea is
2x^2 = bx
what b makes both sides equal (for all x)? b is a number, and no matter what number we can never get this to work for all x. for example
2x^2≠ 2x if x=3 (2*9≠2*3) the only way to get it to work is if we started with
2x^2= bx^2 (x^2 on both sides)

Answer 21

so if we started with
2x^2 + 3x= ax^2 +bx
the only way to make both sides the same is have a=2 and b=3
we match coefficients

Answer 22

it makes a little sense.

Answer 23

First, finish off the problem, then think about it later (sometimes these things make sense after a rest)

x^2+bx= x^2 +2px +p^2+q

b= 2p

an p^2+q= 0
we can think of the left side as x^2+bx+0*x^0
and the right side as
x^2 +2px +(p^2+q)x^0 (of course x^0 is 1)

Answer 24

p in terms of b, we get from
b=2p
solve for p: p= b/2

now use
p^2+q=0
q= -p^2
replace p with b/2 to get q in terms of b

Answer 25

Maybe your intuition will work if you think of each power of x (x^2 , x^1, x^0) as a different animal, and you are told alice and bob have the same number and types of animals: alice has 2 cows and 3 horses. bob has a cows and b horses. what is a and b?
now replace cows with x^2 and horses with x^1
same idea