Inequation with 2 variables - how to proceed?

Question

Find the highest value of $K$ such that $$x^2 - 10x + 40 \ge K$$ $ \forall x \in R$.

Step 1: I'll assume $K$ is equal to $0$ just to make it simpler (Is it OK to do so?) $$x^2 - 10x + 40 \ge 0$$

Step 2: Solving this with the quadratic equation formula:

$$ x = \frac{-(-10) \pm \sqrt[2]{100 - (4)(1)(40)}}{2(1)}$$

Step 3: How to proceed with the negative square root?

$$x = \frac{-(-10) \pm \sqrt[2]{-60}}{2(1)}$$

Possible answers:

$a) 4$ $b) 5$ $c) 6$

$d) 7$ $e) 8$

Answer 1

HINT: Think about the graph of $f(x) = x^2−10x+40$. Does it attain a minimum value? Can you figure out where?

Answer 2

You have

$$x^2 - 10x + 40 \ge K$$

which means the term on the left serves as an upper bound for $K$. To find the biggest value for $K$ is the same as finding the smallest value of that bound.

Minimum via derivative

$$\frac{d}{dx}\left(x^2 - 10x + 40\right) = 0 = 2x-10$$

As it is a simple parabola, I omit the proof that it is a minimum (and not a maximum) Please comment if you need help with that.

The term has a minimum at $x=5$, inserting that into the original formula yields $$5^2 - 10\cdot 5 + 40 \ge K$$ which gives $$15 \ge K$$

15 is the highest value of $K$

Answer 3

Hint: We have that $$x^2-10x = (x-5)^2 - 25. $$ Can you see why? It is due to completing the square. Hence we have that

$$\color{blue}{x^2-10x}+40 = \color{blue}{(x-5)^2 -25} + 40 = (x-5)^2+15. $$

Now recall that the square of a real number cannot be less than $0$. How can you use this?

Answer 4

Hint: here is a way of doing a similar question.

$$4y^2-36y+100\ge K$$

We complete the square to obtain $$(2y-9)^2+19\ge K$$

And we know that squares are always greater than or equal to zero.

Where does the $19$ come from. Well $$19=100-9^2$$ and we can also compute $$b^2-4ac=36^2-4\times 4\times 100=-304=-16\times 19=(2a)^2\times 19$$ so that $\sqrt {b^2-4ac}=4\sqrt{-19}=2a\sqrt {-19}$ and using the quadratic formula to find the roots will cancel the factor $2a$. Notice that $36^2-4\times 4 \times 100=16\times (9^2-100)$. This is not a coincidence.

The negative number under the square root (the negative discriminant $b^2-4ac$) indicates that there are no real roots to the quadratic, so it always stays positive (or negative - this one is positive), and $K\gt 0$.

So the answer is related to the method you tried to use, and if you carry it through algebraically you can explore how it all fits together for yourself. I find that completing the square directly is often the easiest way to solve questions like this without making mistakes.