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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:

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

$d) 7$ $e) 8$

ogondiaz
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4 Answers4

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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?

Eff
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  • So the max reduction, is like you say, $(x - 5)^2 + 14 \ge K$ I could try to guess a number to satisfy this, but the $K$ confuses me. – ogondiaz Sep 08 '15 at 21:34
  • I tried to guide you to find the $K$ with my hint :-). Since a real square can be $0$ - but not less - we have that $(x-5)^2\geq 0$, hence $(x-5)^2 +15 \geq 0 + 15 = 15.$ Therefore the correct answer is $\fbox{K = 15}$. By the way, as some extra info for the curious person, this also reveals that the left hand side achieves its minimum when $(x-5)^2 = 0$, i.e. when $ x = 5$. – Eff Sep 08 '15 at 21:41
  • I see, thanks. So $ 15 \ge K$. But the answer is the lowest value of $k$ is $15$ and the maximus is: $infinity +$? – ogondiaz Sep 08 '15 at 21:44
  • No, $K = 15$ is the highest possible value that satisfies $x^2-10x+40 \geq K$ for all real $x$. There are many, infact an infinite number of, lower values that satisfy the inequality because $$(x-5)^2+15\geq 15 \geq 10 \geq 0 \geq -1000 \geq ... $$ et cetera, et cetera. – Eff Sep 08 '15 at 21:47
  • thanks, I follow your logic, but $15$ is not one of the possible answers. Please, see update. – ogondiaz Sep 08 '15 at 21:58
  • @user3352567 Then there is a mistake in the exercise. The correct answer is $K = 15$. – Eff Sep 09 '15 at 08:15
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HINT: Think about the graph of $f(x) = x^2−10x+40$. Does it attain a minimum value? Can you figure out where?

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    @imranfat He's looking for the highest value of a number which is always lower than or equal to the quadratic. I.e. the minimum. –  Sep 08 '15 at 20:54
  • @imranfat Yes, but the highest value of K is the first integer below the minimum of the function – Feyre Sep 08 '15 at 20:54
  • ok, makes sense – imranfat Sep 08 '15 at 20:55
  • @Bye_World, I don't know about graphs and functions. This is an exercise for my sister that is in 4 grade high school. Please, explain a little bit more. – ogondiaz Sep 08 '15 at 20:58
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    I don't mean any offense but if you don't know what a function is or how to graph one, I think you might not have the skills to solve this problem. –  Sep 08 '15 at 21:00
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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 $K$

null
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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.

Mark Bennet
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