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I'm completely blown away by the difficulty of Spivak. I've managed to work through the first 3 problems, but I feel I'm missing something important to solve these basic inequalities in his 4th problem:

$x^2 + x + 1 > 2$

&

$x^2 + x + 1 > 0$

any suggestions?

  • 0
    Have you tried graphing?2012-06-10

4 Answers 4

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Complete the square: $ x^2 + x + 1 = (x^2 + x + \tfrac 1 4) + \frac 3 4 = \left(x + \frac 1 2 \right)^2 + \frac 3 4. $ That gets you the second one.

For the first one, put everything on one side of the inequality and $0$ on the other side, and procede similarly.

Later addendum in response to vitno's question in the comments below: In general, the process of completing the square looks like this: $ \begin{align} ax^2 + bx + c & = a\left(x^2 + \frac b a x\right) + c \\[12pt] & = a\left(x^2 + \frac b a x + \frac{b^2}{4a^2}\right) + c - a\left(\frac{b^2}{4a^2}\right) \tag{$\begin{array}{c} \text{completing} \\ \text{the square}\end{array}$} \\[12pt] & = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}. \end{align} $ Say you have a particular case: $ 3x^2 + 20 x + 7. $ Proceed as follows: $ 3\left(x^2 + \frac{20}{3} x \right) + 7. $ Take half the coefficient of the first-degree term and square it, getting $(10/3)^2$. Add this in the appropriate place, and substract it out later: $ 3\underbrace{\left(x^2 + \frac{20}{3} x + \left(\frac{10}{3}\right)^2 \right)}_{\text{a perfect square}} + 7 - 3\left(\frac{10}{3}\right)^2 $ $ = 3\left(x + \frac{10}{3}\right)^2 - \frac{79}{3}. $

Knowing how and when to complete the square is useful.

Remember this: The purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term.

  • 0
    For a quadratic expression, you get the coefficient of $x^2$ to be 1, by dividing through if necessary. If the coefficient of $x$ is then $b$, you get the expression to be $(x+\frac b2)^2 + C $, where C is a constant term independent of $x$.2012-06-10
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$x^2+x+1>2 \iff x^2+x+\frac{1}{4}>\frac{5}{4}$

Thus, $(x+\frac{1}{2})^2\gt\frac{5}{4}$. We can reduce this to $|x+\frac{1}{2}|>\frac{\sqrt{5}}{2}$. Thus, our final answer will be

$x<-\frac{\sqrt{5}+1}{2}$ or $x>\frac{\sqrt{5}-1}{2}$.

For the second answer, notice that we can apply the same strategy to the previous problem and get $(x+\frac{1}{2})^2>-\frac{3}{4}$. Notice, that if $x$ is a real number, then squaring $(x+\frac{1}{2})$ will never create a negative number, so this inequality holds for all real numbers.

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Hint: start by considering the graph of $y = x^2+x+1$. What is its shape? Where is $y=0$ or $2$?

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We have $x^2+x+1=(x+1/2)^2+3/4$. Now any inequalities you want to prove involve only a square.

So for example for $x^2+x+1 \gt 2$, rewrite as $(x+1/2)^2\gt 5/4$, giving $|x+1/2|\gt (1/2)\sqrt{5}$, which can be rewritten in various ways.

For $x^2+x+1 \gt 0$, note that in fact $x^2+x+1 \ge 3/4$, since any square is $\ge 0$.