Quadratic Inequalities: Notion quadratic inequalities
Quadratic inequalities
Quadratic inequalites are inequalities where both sides of the equal sign contain polynomials in a variable that are constant, linear or quadratic.
Examples are #5x-2 \gt x^2+4# and #t^2-3t+10 \le 5t-2#.
If there are no quadratic terms in the inequality, then it is a linear inequality. For example: #-3t+10\le 5t-2#.
If there are not any linear terms either, then the inequality is a statement about two numbers that says which of the thow number is bigger than (or equal to) the other. For example: #10\le -2#. This statement does not have to be true.
- Rewrite the inequality to the form #ax^2+bx+c \gt 0#.
- Replace the inequality by an equation and solve this equation.
- Find out (for example with the help of a graph) for which #x# the function #ax^2+bx+c # is positive.
The first step transforms the inequality in a standard form.
The second step is needed to investigate when the function #ax^2+bx+c# switches signs.
If #ax^2+bx+c=0# has no solutions, then this never happens. The function is always positive or always negative. The answer can then be formulated as "all" or "none".
If #ax^2+bx+c=0# has one solution, then this solution is #x=-\dfrac{b}{2a}# and the sign in that point is zero; the function is always positive or always negative, except in #x=-\dfrac{b}{2a}#. We than formulate the answer as "none" or #x\lt -\dfrac{b}{2a}\vee x\gt -\dfrac{b}{2a}#.
If #ax^2+bx+c=0# has two solutions, say #x_0# and #x_1#, with #x_0\lt x_1#, then the sign on the interval #(x_0,x_1)# differs from the sign outside of this interval. In the case where #a\gt0# the graph of the function is a parabola opening upward and the function is exactly positive if #x\lt x_0# or #x\gt x_1#. The answer is formulated as #x\lt x_0\vee x\gt x_1#. In the case where #a\lt0# the graph of the function is a parabola opening downward and the function is positive if and only if #x\gt x_0# and #x\lt x_1#. Hence the answer can be formulated as #x\gt x_0\wedge x\lt x_1#.
Solve #x^2-5\cdot x-6\lt 2\cdot x^2-27\cdot x+114#.
Provide your answer in the form #x = a# or #(x\lt a) \lor (x\gt b)#, or similar. You can use #\vee# and #\wedge#.
The first step is rewriting the inequality in such a way that the right hand side is equal to #0#. This is done by subtracting all terms present in the right hand side from both the right and left hand side. The result is #-x^2+22\cdot x-120\lt 0#.
Now we solve the equation #-x^2+22\cdot x-120=0# with unknown #x#. The abc-formula gives #x=10\vee x=12#.
According to the theory of Quadratic functions, the graph of the function #-x^2+22\cdot x-120# is a downward parabola. Hence, within the interval #[10,12]#, the values of #-x^2+22\cdot x-120# will be positive. The conclusion is that #x# is a solution to the inequality if and only if #x\lt 10\vee x\gt 12#.
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