vrajesh wrote:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 f[highlight]or all values of x[/highlight], what is the value of c?
(1) d = 3
(2) b = 6
since expanding (x + d)^2 = x^2 + 2xd + D^2
hence we have
bx + c = 2xd + D^2
we need to know value of b and value of d to get the correct answer.
Can you please explain if it is other wise.
Given \(x^2 + bx + c = (x+d)^2\)
Expanding on both sides, we get: \(x^2 + 2dx + d^2 = x^2 + bx + c.\) We can cancel the x^2 on both sides and that leaves us with: \(2dx + d^2 = bx + c\)
And this is valid for all values of x, we are given. Let's just substitute two values of x:
\(x = 1\)
Then \(2d + d^2 = b + c\)
\(x = 2\)
\(4d + d^2 = 2b + c\)
Now taking these two equations together:
\(2d + d^2 = b + c\\
4d + d^2 = 2b + c\)
Multiplying the first equation by 2 and solving, we get:
\(4d + 2d^2 = 2b + 2c\\
4d + d^2 = 2b + c\)
So we get: \(d^2 = c\)
Thus, from our two statements, we know that statement one is sufficient by itself. Now look at statement 2. And look at the first equation we had: \(2d + d^2 = b + c.\) But then \(c = d^2\), which means that \(2d = b\). So, if b is given, you can find d and hence c. Thus statement 2 is also sufficient.
Thus the trick here is to read the question carefully. If it's valid for all values of x, it's valid for any two values of x. You can arbitrarily pick x. In fact, picking x = 0 might be even better and help you solve the problem much faster, in hindsight.