Question

4x^2-35x+49

Answer 1

That's a lovely expression. What is to be done.

By the way, I am immediately tantalized by the fact that the first and last terms are perfect squares. I wonder if that will help us?

Answer 2

I have no clue how to even start solving this problem. Do you know how? Can you show me the steps?

Answer 3

Well, so far we have only an expression. There are NO instructions?

Answer 4

Nope. There isn't any instructions

Answer 5

Wow, that's really sad. Why in the name of reason are you given nothing? Utterly senseless.

Anyway, perhaps we are to factor it?

4x^2-35x+49

Like I said up top, since both the first and last terms are perfect squares, I'm tempted just to give it a shot without any further effort.

4x^2 = (2x)^2
49 = (7)^2

Does this work? (2x-7)^2

Multiply that out and see if we get back where we started/.

Answer 6

I am so sorry. I Honestly have no idea what I'm doing...

Answer 7

\[ax^2+bx+c=0\]
\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[(x-x_1)(x-x_2)=0\]

Answer 8

Have you ever multiplied to binomials?

(a+2)(a-3) = a^2 +2a - 3a - 6 = a^2 -a -6

Something like that?

Answer 9

Nope, I have not

Answer 10

Fair enough. Then you have no immediate capacity to solve what we have imagined this problem to be. If this is a placement exam, the course where this material is covered may be the right course for you. If you are already in this class, you are in for a very unfortunate semester. A quick chat with your academic adviser may be the best course of action. Good luck.