Prove that if $x$ is a real number in the range $12 - 7x + x^2 \ge 0.$
Which type of proof should I use to solve this? At first I thought direct proof. Choosing a number between $0$ and $3$ and attempting to solve?
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Hint: look at the roots of the quadratic. If your quadratic has roots $r_1 < r_2$ note that it is nonnegative for $x \le r_1$ and $x \ge r_2$. – sardoj Apr 28 '15 at 21:22
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Hints:
$$12-7x+x^2=\left(x-\frac{7}{2}\right)^2+12-\frac{49}{4}=\left(x-\frac{7}{2}\right)^2-\frac{1}{4}$$
$$0\leq x\leq 3\implies \frac{1}{4}\leq \left(x-\frac{7}{2}\right)^2\leq \frac{49}{4}$$
Prasun Biswas
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1It's easy to factorize the quadratic, draw a "Why so serious" on a straight line. – GohP.iHan Apr 30 '15 at 03:11
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1Haha, this ain't a troll post. My solution is easier: draw a smiling face because 3,4 are roots. and voila, the quadratic is nonnegative for $0\leq x\leq 3$ – GohP.iHan Apr 30 '15 at 04:51
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@GohP.iHan, yeah, I saw that now. It's really a nice way to look at it. :) – Prasun Biswas Apr 30 '15 at 04:56
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Evaluate the endpoints. What happens when $x = 0$, $x=3$? Can the expression ever be negative? There would have to be a point where is moves from the positive side to the negative side, i.e, where the expression is zero. Compute where that happens, or show that it can't, using the quadratic formula.
user3697176
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