If 4x^2+20x+r=(2x+s)^2 for all values of x, what is the value of r-s?
Can you please help me?
just expand and equate coefficients
4x^2+20x+r = 4x^2+4sx+s^2
4s=20
s=5
r = s^2 = 25
r-s = 20
Of course, I can help you with that!
To find the value of r-s, we need to compare the given equation 4x^2+20x+r=(2x+s)^2 with the general form of a perfect square trinomial: (a+b)^2.
When we expand (2x+s)^2, we get 4x^2 + 4xs + s^2.
Comparing the two equations, we can see that the coefficients of the x^2 terms are the same (both are 4x^2), the coefficients of the x terms are the same (both are 20x), and the constant terms are also the same (both are r).
However, the coefficient of the xs term in the second equation is 4s, which does not match the coefficient of the xs term in the first equation, which is 20x.
Therefore, for the given equation 4x^2+20x+r=(2x+s)^2 to be true for all values of x, the coefficient of the xs term in the second equation must also be 20x.
Therefore, we can equate 4s to 20x and solve for s:
4s = 20x
s = 5x
Now we can find the value of r-s by substituting the value of s back into the original equation and comparing coefficients:
r - s = r - 5x
Since there is no indication of any particular value for x in the given equation, we cannot solve for r - s without additional information.