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# Solving Quadratic Inequalities: Graphic Approach

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Joined: 02 Sep 2009
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23 Apr 2014, 10:57
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Say we need to find the ranges of $$x$$ for $$x^2-4x+3<0$$. $$x^2-4x+3=0$$ is the graph of a parabola and it look likes this:

Intersection points are the roots of the equation $$x^2-4x+3=0$$, which are $$x_1=1$$ and $$x_2=3$$. "<" sign means in which range of $$x$$ the graph is below x-axis. Answer is $$1<x<3$$ (between the roots).

If the sign were ">": $$x^2-4x+3>0$$. First find the roots ($$x_1=1$$ and $$x_2=3$$). ">" sign means in which range of $$x$$ the graph is above x-axis. Answer is $$x<1$$ and $$x>3$$ (to the left of the smaller root and to the right of the bigger root).

This approach works for any quadratic inequality. For example: $$-x^2-x+12>0$$, first rewrite this as $$x^2+x-12<0$$ (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).

$$x^2+x-12<0$$. Roots are $$x_1=-4$$ and $$x_1=3$$ --> below ("<") the x-axis is the range for $$-4<x<3$$ (between the roots).

Again if it were $$x^2+x-12>0$$, then the answer would be $$x<-4$$ and $$x>3$$ (to the left of the smaller root and to the right of the bigger root).

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28 May 2014, 03:23
Check out all important topics we have in Main Math forum HERE.

Important topics in PS forums.

Important topics in DS forums.
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16 Aug 2018, 19:19
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Re: Solving Quadratic Inequalities: Graphic Approach &nbs [#permalink] 16 Aug 2018, 19:19
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