w^2+2w+d is a perfect square?

Question

find the value of d such that

w^2+2w+d

is a perfect square

Answer 1

If you had:
(w + 1)²

That would become:
w² + 2w + 1

Comparing that to your equation, it's obvious that d = 1.


When you have something in the form:
x² + bx + d

It will be a perfect square when:
d = (b/2)²

This means take half the coefficient on the x-term and square it.

In your example:
d = (2/2)²
d = 1

Answer 2

d=1

Answer 3

a perfect square can be an algebraic expression that can be factored as the square of some other expression, e.g. a^2 ± 2ab + b^2 = (a ± b)^2

so for w^2+2w+d to be a perfect square, d must be a number which would make the expression equal to (w+d)^2 or (w-d)^2

then d must equal to 1 and so we will have (w+1)^2=w^2+2w+1

Answer 4

w^2+2w+d
if d=1
w^2+2w+1=(w+1)^2

Answer 5

w^2+2w+d=w(w+2+d/w)=x^2

i.e X=a non defined value.

there fore d=1

Answer 6

[a+b]^2=a^2+2ab+b^2
here we have
2ab=2w
b=1
d=b^2=1^2
=1