Problem 50
Question
Give graphical and algebraic explanations of why \(x^{2}+1<-x\) has no real- valued solution.
Step-by-Step Solution
Verified Answer
The inequality \(x^{2}+1<-x\) has no real-valued solution because the rearranged form of the inequality, \(x^{2} + x + 1 > 0\), corresponds to a parabola that opens upwards. This parabola never intersects or goes below the x-axis, meaning that all real values of \(x\) result in \(x^{2} + x + 1 > 0\). As we were interested in when this quantity is less than 0, and no such values of \(x\) exist, we can conclude that there are no real-valued solutions.
1Step 1: Rearrange the inequality
Rearrange the inequality \(x^{2} + 1 < -x\) to \(x^{2} + x + 1 > 0\). This is an easier form to work with, as we can now equate this to a function in the form \(y = x^{2} + x + 1\), which we can graph and analyze.
2Step 2: Plot the function
Plot the function \(y = x^{2} + x + 1\). This is a parabola that opens upwards. The vertex of this parabola is given by \(-b/{2a} = -1/2\). As the parabola opens upwards, this vertex is the lowest point of the graph, representing the smallest value \(y\) can take.
3Step 3: Determine solutions to the inequality
Looking at the graph, it is clear that the parabola never intersects or goes below the x-axis (where \(y = 0\)). The smallest value it ever takes is when \(x = -1/2\), and at this point, \(y = 1 -1/2 + 1 = 3/4\), which is greater than 0. Therefore, all real values of \(x\) make \(x^{2} + x + 1 > 0\). Since we were interested in when this quantity is less than 0, and no such values of \(x\) exist, we can conclude that there are no real-valued solutions to the inequality \(x^{2}+1<-x\).
Key Concepts
Graphical AnalysisInequality SolutionsParabola
Graphical Analysis
Graphical analysis is a powerful tool when solving quadratic inequalities like \(x^{2} + 1 < -x\). By transforming this inequality into the form of \(x^{2} + x + 1 > 0\), we can graph the function \(y = x^{2} + x + 1\).
The graph of this function is a parabola that opens upwards. This is indicative of a positive quadratic coefficient. The vertex, which is the turning point of the parabola, can be found using the formula \(-\frac{b}{2a}\), which yields \(-\frac{1}{2}\) for this case.
Once we have the graph in place, we can observe that the parabola does not touch or pass below the x-axis (where \(y = 0\)). Thus, every point on the parabola is above the x-axis, reinforcing that the smallest value the function can take is \(\frac{3}{4}\) at \(x = -\frac{1}{2}\), confirming it is always greater than zero for all real \(x\). This graphical method visually proves that no real solutions exist for the original inequality \(x^{2} + 1 < -x\).
The graph of this function is a parabola that opens upwards. This is indicative of a positive quadratic coefficient. The vertex, which is the turning point of the parabola, can be found using the formula \(-\frac{b}{2a}\), which yields \(-\frac{1}{2}\) for this case.
Once we have the graph in place, we can observe that the parabola does not touch or pass below the x-axis (where \(y = 0\)). Thus, every point on the parabola is above the x-axis, reinforcing that the smallest value the function can take is \(\frac{3}{4}\) at \(x = -\frac{1}{2}\), confirming it is always greater than zero for all real \(x\). This graphical method visually proves that no real solutions exist for the original inequality \(x^{2} + 1 < -x\).
Inequality Solutions
To solve quadratic inequalities, we perform both algebraic manipulations and graphical inspections. Our initial step took \(x^{2} + 1 < -x\) and rearranged it to \(x^{2} + x + 1 > 0\).
This transformation allows us to visualize the problem as finding where our quadratic function \(y = x^{2} + x + 1\) has positive values. Algebraically, solving such inequalities would often involve factoring or identifying zeros where the expression is equal to zero. However, if the graph shows that the quadratic expression is never zero or negative, like in our case, it indicates there are no real solutions to meet the inequality condition.
In sum, solving inequalities using graphical and algebraic means offers comprehensive insights. Here, observing that the quadratic is always positive confirms the absence of any real solutions, as was shown with \(x^{2} + 1 < -x\).
This transformation allows us to visualize the problem as finding where our quadratic function \(y = x^{2} + x + 1\) has positive values. Algebraically, solving such inequalities would often involve factoring or identifying zeros where the expression is equal to zero. However, if the graph shows that the quadratic expression is never zero or negative, like in our case, it indicates there are no real solutions to meet the inequality condition.
In sum, solving inequalities using graphical and algebraic means offers comprehensive insights. Here, observing that the quadratic is always positive confirms the absence of any real solutions, as was shown with \(x^{2} + 1 < -x\).
Parabola
A parabola is a U-shaped curve that represents the graph of a quadratic function such as \(y = x^{2} + x + 1\). The general form of a quadratic is \(ax^{2} + bx + c\), where \(a\), \(b\), and \(c\) are constants. The sign of \(a\) determines whether the parabola opens upwards or downwards; a positive \(a\) opens upwards, while a negative \(a\) opens downwards.
The vertex formula \(-\frac{b}{2a}\) provides the x-coordinate of the vertex of the parabola, which is the minimum or maximum point of the curve, depending on its direction.
In our example, \(a = 1\), \(b = 1\), and \(c = 1\), resulting in a minimum point at \(x = -\frac{1}{2}\). This parabola does not intersect the x-axis since its vertex is above it and the arms extend upwards, indicating it is always positive for real number inputs. This conclusion helps substantiate that for the inequality \(x^{2} + 1 < -x\), no real solutions exist because \(y = x^{2} + x + 1\) is never less than zero.
The vertex formula \(-\frac{b}{2a}\) provides the x-coordinate of the vertex of the parabola, which is the minimum or maximum point of the curve, depending on its direction.
In our example, \(a = 1\), \(b = 1\), and \(c = 1\), resulting in a minimum point at \(x = -\frac{1}{2}\). This parabola does not intersect the x-axis since its vertex is above it and the arms extend upwards, indicating it is always positive for real number inputs. This conclusion helps substantiate that for the inequality \(x^{2} + 1 < -x\), no real solutions exist because \(y = x^{2} + x + 1\) is never less than zero.
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