<p>2007 PSAT (Saturday) : Section 2: Question 16</p>
<p>16) x^2+kx+4=(x+c)^2</p>
<p>In the equation above, k and c are positive constants. If the equation is true for all values of x, what is the value of k?</p>
<p>choices:
A) 2
B) 4
C) 8
D) 16
E) 32</p>
<p>Answer: B, 4</p>
<p>i hate these types of questions, can anyone explain me how you get 4? i got 2.
thnxss</p>
<p>We have x^2+kx+4=(x+c)^2
Foil out (x+c)^2 and we'll get
x^2+2cx+x^2
Therefore c^2 = 4 because the y intercept = that of the first eqn.
Thus c = 2.
x^2+2(2)X+4 = x^2+kx+4
K=4.</p>
<p>x^2+kx+4=(x+c)^2
x^2 + kx + 4 = x^2 + 2cx + c^2</p>
<p>Therefore
k = 2c and c^2 = 4
c = +- 2 but we disregard the - because it's a positive constant</p>
<p>k = 2c
k = 2(2)
k = 4</p>
<p>Here's a shortcut.
(x+c)^2 suggests that it is a perfect square.</p>
<p>With perfect squares, there is a cool rule: sqrt(c) x 2= B</p>
<p>sqrt[4]=2; 2x2 = 4.</p>