Quadratic equations: Quadratic inequalities
Quadratic inequalities
Just as with linear inequalities, we can create an inequality with quadratics. We will first take a look at how to solve a quadratic inequality.
We solve the inequality #\blue{2x^2+4x+3} \gt \green{x^22x+6}#.
Step 1 
We first solve the equality #\blue{2x^2+4x+3} = \green{x^22x+6}#. Therefore we first reduce the equality to #0#. \[\begin{array}{rcl}2x^2+4x+3&=&x^22x+6 \\ &&\phantom{xx}\blue{\text{the equation to solve}}\\ 3x^2+6x3 &=&0 \\ &&\phantom{xx}\blue{\text{reduce to }0} \end{array}\] Next we apply the quadratic formula. Therefore we define the letters #a#, #b# and #c#. \[a=3, b=6 \text{ and } c=3\] Next, we calculate the discriminant. \[D=6^24 \cdot 3 \cdot 3 =72\] After that, we determine the solutions \[x=\frac{6\sqrt{72}}{2\cdot 3} \lor x=\frac{6+\sqrt{72}}{2\cdot 3} \] Which we simplify to: \[x=1 \sqrt{2} \lor x=1+ \sqrt{2} \] 
Step 2 
We create the graphs of #y=\blue{2x^2+4x+3}# and #y=\green{x^22x+6}\ (dashed)#. The intersection points are drawn in orange. 
Step 3 
We will determine the solution by means of step 1 and 2. On the left of #x=1\sqrt{2}# and on the right of #x=1+\sqrt{2}# the graph of #y=\blue{2x^2+4x+3}# lies above the graph of #y=\green{x^22x+6}#. hence, the solution is #x \lt 1\sqrt{2} \lor x \gt 1+\sqrt{2}#. 
In general, we can apply the following procedure.
Solving a quadratic inequality
Procedure  
We solve the inequality #\blue{a_1 x^2+b_1x+c_1} \gt \green{a_2x^2+b_2x+c_2}\ (dashed)#, in which #a_1 \ne 0#. 


Step 1  We first solve the equality \[\blue{a_1 x^2+b_1x+c_1} = \green{a_2x^2+b_2x+c_2}\]  
Step 2  We draw the graphs #y=\blue{a_1 x^2+b_1x+c_1}# and #y=\green{a_2x^2+b_2x+c_2}#.  
Step 3  With help of step 1 and 2, determine for which value of #x# the inequality holds. In the coordinate system, the bigger graph is the one that lies above the other. 
Note that this procedure also holds for the inequality signs #\geq#, but the #x#values of the intersection points are also part of the solution.
Step 1  We solve the equality #4\cdot t^2+3\cdot t2=t^2+4\cdot t10#. This is done in the following way: \[\begin{array}{rcl} 4\cdot t^2+3\cdot t2 &=& t^2+4\cdot t10\\ &&\phantom{xxx}\blue{\text{the original equality}}\\ 5\cdot t^2t+8&=&0\\ &&\phantom{xxx}\blue{\text{all terms moved to the left hand side}}\\ \text{discriminant } &=& b^24ac \\ &&\phantom{xxx}\blue{\text{formula discriminant}}\\ &=& \left(1\right)^2  4 \cdot \left(5\right) \cdot 8 \\&&\phantom{xxx}\blue{\text{formula entered}}\\ &=& 161\\&&\phantom{xxx}\blue{\text{calculated}}\\ \text{number of solutions } &=& 2\\&&\phantom{xxx}\blue{\text{since discrimant bigger than }0}\\ t=\dfrac{b\sqrt{D}}{2 \cdot a} &\lor& t=\dfrac{b+\sqrt{D}}{2 \cdot a} \\&&\phantom{xxx}\blue{\text{formula solutions}}\\ t=\dfrac{\left(1\right)  \sqrt{161}}{2 \cdot \left(5\right)} &\lor& t=\dfrac{\left(1\right) + \sqrt{161}}{2 \cdot \left(5\right)} \\&&\phantom{xxx}\blue{\text{formula entered}}\\ \displaystyle t= {{\sqrt{161}1}\over{10}} &\lor& \displaystyle t = {{\sqrt{161}1}\over{10}} \\&&\phantom{xxx}\blue{\text{calculated}}\\ \end{array}\] 
Step 2  We sketch the graphs of #y=4\cdot t^2+3\cdot t2# (blue solid) and #y=t^2+4\cdot t10# (green dashed). 
Step 3  We can now read the answer of the inequality. \[t\lt {{\sqrt{161}1}\over{10}}\lor t\gt {{\sqrt{161}1}\over{10}}\] 
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