Quadratic Inequalities

So, we will factorise this quadratic, and then use what would be the solutions to help us plot the graph. Observing that $(-2)times(-3)=6$ and $(-2)+(-3)=-5$, we get that inequality can be written as,

$begin{aligned}x^2-5x+6&leq0 (x-2)(x-3)&leq0end{aligned}$

So, the roots of this quadratic would be $x=2$ and $x=3$, therefore the graph looks like

The inequality in the question is $x^2-5x+6leq0$, so the question is: when does the graph go below zero? We can see that it goes below zero when $x$ is between $2$ and $3$. Therefore, the solution is,

$2leq xleq3$

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So, we will factorise this quadratic, and then use what would be the solutions to help us plot the graph.

Taking $x$ out as a factor, we get

$begin{aligned}x^2-3x&>0 x(x-3)&>0end{aligned}$

So, the roots of this quadratic would be $x=0$ and $x=3$, therefore the graph looks like

The inequality in the question is $x^2-3x>0$, so the question is: when does the graph go above zero? We can see that it goes above zero when $x$ is less than $0$ and also when $x$ is above 3. Therefore, the solution is,

$x<0,text{ or },x>3$

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We solve this inequality by simply rearranging it to make $p$ the subject,

 $begin{aligned}3p^2+8 &> 20 3p^2 &> 12  p^2&>4 end{aligned}$

Hence $p$ can take any value greater than $2$,  $p>2$ or less than $-2$, $p<-2$, therefore the graph looks like,

The inequality in the question is $3p^2-12>0$, so the question is: when does the graph go above zero? We can see that it goes above zero when $p$ is less than $-2$ and also when $p$ is above $2$. Therefore, the solution is,

$p<-2,text{ or },p>2$

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We solve this inequality by simply rearranging it to make $x$ the subject,

 $begin{aligned}x^2-3x+2&<0 (x-2)(x-1)&<0  end{aligned}$

Hence $x$ can take any value between  $1<x<2$

The inequality in the question is $x^2-3x+2<0$, so the question is: when does the graph go below zero? We can see that it goes below zero between the values $x$ is greater than 1 and $x$ is less than  2. Therefore, the solution is,

$1<x<2$

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We solve this inequality by simply rearranging it to make $x$ the subject,

 $begin{aligned}x^2-5x+24&<5x+8 x^2-10x+16&<0 (x-8)(x-2)&<0end{aligned}$

Hence $x$ can take any value between  $2<x<8$

The inequality in the question is $x^2-10x+16<0$, so the question is: when does the graph go below zero? We can see that it goes below zero between the values $x$ is greater than $2$ and $x$ is less than  $8$. Therefore, the solution is,

$2<x<8$

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