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Solve the inequality $4t^2 \le 9t - 2.$

 Jun 25, 2021
 #1
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$  4t^2\le 9t-2 \ \ \ \Rightarrow \ \ \  4t^2+2\le 9t  \ \ \ \Rightarrow \ \ \   4t^2+2-9t\le \:0 $

 

 

a trinomial

 

$a=4$   ;   $b=-9 $   ;    $c=2$

 

plug that into $ _2t_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a}  $:

 

$_2t_1=\frac{-\left(-9\right)\pm \sqrt{\left(-9\right)^2-4\cdot \:4\cdot \:2}}{2\cdot \:4}$

 

$_2t_1=\frac{-\left(-9\right)\pm \sqrt{81-32}}{2\cdot \:4}$

 

$_2t_1=  =\frac{-\left(-9\right)\pm \:7}{2\cdot \:4}$

 

$t_1=\frac{9+7}{2\cdot \:4}$   and   $t_2=\frac{9-7}{2\cdot \:4}$

 

those, respectively give you $   t_1=2  $   and $   t_2=\frac{1}{4} $

 

thus, we get $  \boxed{\frac{1}{4}\le t\le 2}  $

 Jun 25, 2021

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