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I want to ask you something about the following quadratic inequality: $$9x^2+12x+4\le 0.$$ Let us find the "=0" points. Here we have $9x^2+12x+4=9(x+\frac{2}{3})^2\ge0$ (we can factor this by calculating the discriminant $D=36-36=0$ and using $ax^{2}+bx+c=a(x-x_1)(x-x_2)$). I am practising the wavy curve method (the method of intervals) to solve quadratic inequalities, so I would want to solve it using this way. What is the general method to approach the quadratic inequality when the discriminant is zero, and we want to use the wavy curve method? enter image description here

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  • $\begingroup$ Imagining the curve, it has a double touch point when $x=-2/3$ and it is a positive quadratic so the only point at which $f(x)\le0$ is when $x=-2/3$ and $f(x)=0$. $\endgroup$ – Peter Foreman Apr 2 '20 at 10:32
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$$(3x+2)^2\leq0$$ gives $$3x+2=0$$ or $$x=-\frac{2}{3}.$$ Also, if you want to use $\Delta$, so for $\Delta=0$ we have the following formula for the root.

$$x_1=-\frac{b}{2a}.$$ We obtain: $$x_1=-\frac{12}{2\cdot9}=-\frac{2}{3}.$$

There are four cases:

  1. $(x-a)^2\geq0.$ We obtain $\mathbb R$;

  2. $(x-a)^2>0.$ we obtain $(-\infty,a)\cup(a,+\infty)$;

  3. $(x-a)^2\leq0$. We obtain $\{a\}$ and

  4. $(x-a)^2<0$. We obtain $\oslash$

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  • $\begingroup$ Thank you for the response! Our teacher told us always to plot the roots on the number line and determine the sign in each interval (the roots divide the number line in different intervals). How should it look in this case? How to represent $-\dfrac{2}{3}$ and the intervals with the signs? $\endgroup$ – Knowledge Greedy Apr 2 '20 at 10:38
  • $\begingroup$ @Knowledge Greedy Just take $a$ on the $x$-axis and see the sign of the expression. I don't know to draw in the net. Sorry. $\endgroup$ – Michael Rozenberg Apr 2 '20 at 10:41
  • $\begingroup$ What type of dot should I use when taking $-\dfrac{2}{3}$ on the number line? $\endgroup$ – Knowledge Greedy Apr 2 '20 at 10:42
  • $\begingroup$ @Knowledge Greedy If there is an equality, so you need to use the full point, of course. In our case you take a full point $-\frac{2}{3}.$ $\endgroup$ – Michael Rozenberg Apr 2 '20 at 10:43
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    $\begingroup$ @Knowledge Greedy You need "less or equal", but you got "greater or equal". Id est, you got that only $-\frac{2}{3}$ is valid. Now it's clear? $\endgroup$ – Michael Rozenberg Apr 2 '20 at 10:50
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In this case there is no need to find the intersection with $y=0$, because you re considering the inequality: $$9(x+\frac{2}{3})^2\ge0$$ which is always true for every $x$ in $R$. On the other side, when you have an inequality in the form: $$ax^2+bx+c\geq0$$ you can always rewite it as: $$a(x-x_1)(x-x_2)\geq0$$ and then study the sign of $x-x_1$ and $x-x_2$. The interval where both factors are positive or both are negative are correct and are solutions to the inequality.

Also, I recommend you to know this theorem:

Give a polynomial $P(x)=a^+bx+c$, the sign of $P(x)$ betweehn the roots $(P(x)=0)$ is opposite respect to the sign outsider the interval of the roots.

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  • $\begingroup$ Thank you! I appreciate it! <3 $\endgroup$ – Knowledge Greedy Apr 2 '20 at 10:58
  • $\begingroup$ But there's a high-school theorem for the sign of a quadratic polynomial! It is the sign of the leading coefficient outside the interval of the roots (if any), the opposite sign between the roots. $\endgroup$ – Bernard Apr 2 '20 at 11:25
  • $\begingroup$ @Bernard: of course, there is. I will include it in my answer. $\endgroup$ – Matteo Apr 2 '20 at 11:30

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