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^2-2x+6}#.
Step 1 |
We first solve the equality #\blue{2x^2+4x+3} = \green{-x^2-2x+6}#. Therefore we first reduce the equality to #0#. \[\begin{array}{rcl}2x^2+4x+3&=&-x^2-2x+6 \\ &&\phantom{xx}\blue{\text{the equation to solve}}\\ 3x^2+6x-3 &=&0 \\ &&\phantom{xx}\blue{\text{reduce to }0} \end{array}\] 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. \[D=6^2-4 \cdot 3 \cdot -3 =72\] 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^2-2x+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^2-2x+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#. |
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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 p^2-8\cdot p+9=-2\cdot p^2-4\cdot p+2#. This is done in the following way: \[\begin{array}{rcl} -4\cdot p^2-8\cdot p+9 &=& -2\cdot p^2-4\cdot p+2\\ &&\phantom{xxx}\blue{\text{the original equality}}\\ -2\cdot p^2-4\cdot p+7&=&0\\ &&\phantom{xxx}\blue{\text{all terms moved to the left hand side}}\\ \text{discriminant } &=& b^2-4ac \\ &&\phantom{xxx}\blue{\text{formula discriminant}}\\ &=& \left(-4\right)^2 - 4 \cdot \left(-2\right) \cdot 7 \\&&\phantom{xxx}\blue{\text{formula entered}}\\ &=& 72\\&&\phantom{xxx}\blue{\text{calculated}}\\ \text{number of solutions } &=& 2\\&&\phantom{xxx}\blue{\text{since discrimant bigger than }0}\\ p=\dfrac{-b-\sqrt{D}}{2 \cdot a} &\lor& p=\dfrac{-b+\sqrt{D}}{2 \cdot a} \\&&\phantom{xxx}\blue{\text{formula solutions}}\\ p=\dfrac{-\left(-4\right) - \sqrt{72}}{2 \cdot \left(-2\right)} &\lor& p=\dfrac{-\left(-4\right) + \sqrt{72}}{2 \cdot \left(-2\right)} \\&&\phantom{xxx}\blue{\text{formula entered}}\\ \displaystyle p= {{3\cdot \sqrt{2}-2}\over{2}} &\lor& \displaystyle p = {{-3\cdot \sqrt{2}-2}\over{2}} \\&&\phantom{xxx}\blue{\text{calculated}}\\ \end{array}\] |
Step 2 | We sketch the graphs of #y=-4\cdot p^2-8\cdot p+9# (blue solid) and #y=-2\cdot p^2-4\cdot p+2# (green dashed). |
Step 3 | We can now read the answer of the inequality. \[p\lt {{-3\cdot \sqrt{2}-2}\over{2}}\lor p\gt {{3\cdot \sqrt{2}-2}\over{2}}\] |
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