Given that x is real, prove that x^2 - 4x + 9 is always positive.?

Question

Right, this is obviously Math... Don't worry, this isn't a homework I'm making you do for me. I'm practising for my exams... I know it's supposed to be easy but...yeah. Please?

By the way...I'm not sure if it's important to mention it but this is a GCE 'A' Level H2 sub-question.

Answer 1

You can show that the discriminate (b^2-4ac) is negative and that the function is positive for at least one value of x. That should pretty much do it. If the quadratic has no real roots then it doesn't touch the x-axis.

I don't know what GCE 'A' Level H2 means but I'm guessing the level of 'proof' required isn't all that rigorous.

Okay. Here's a more rigorous proof.

Assume there is a value, k, such that f(k) < 0. We know that f(0) = 9. Since f(x) is a continuous function we know there must exist k1 between k and 0 such that f(k1) = 0. Since the discriminate of f is negative, we know that no such value exists. Therefore, we have a contradiction.

Answer 2

You can prove it by assuming the opposite, and you get a non-real result.

Assume x^2 - 4x + 9 <= 0

As you solve this, you will get to a point where you need the square root of -20, which doesn't exist in the real numbers. Therefore, the equation always has to be positive.

Answer 3

Evan B's proof is sufficient.

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Another way

You could find the critical point (by differentiating):

F(x) = x^2 - 4x + 9
F'(x) = 2x - 4
There is only one critical point, and it is at x = 2

F"(x) = +2 (the second diff. being positive means that the critical point is a minimum)

So, the value of F(x) is never less than the value of F(2).

F(2) = 2^2 - 4*2 + 9 = 4 - 8 + 9 = +5

Therefore, F(x) can never be less than +5 (so it cannot be negative)