Question

Find the solution set of \(\mathrm{x}^{2}-6 \mathrm{x}+10>0\) by the graphical method.

Short answer

The solution set for the inequality \(x^2 - 6x + 10 > 0\) is the entire real number line, as the parabola represented by \(f(x) = x^2 - 6x + 10\) opens upwards and doesn't intersect the x-axis. The vertex of the parabola is at point \(V(3,1)\). The solution set can be written using interval notation as \((-\infty, +\infty)\).

Step-by-step solution

1. Rewrite the inequality in the standard quadratic form

Rewrite the inequality as \(f(x) = x^2 - 6x + 10\). We are looking for the x values for which \(f(x) > 0\).

2. Find the vertex of the parabola

The vertex of a parabola \(y = ax^2 + bx + c\) can be found using the formula \(V(\frac{-b}{2a}, f(\frac{-b}{2a}))\). In our case, \(a = 1, b = -6\), and \(c = 10\). Plug these values into the formula: - Coordinate x: \(\frac{-(-6)}{2(1)} = \frac{6}{2} = 3\) - Coordinate y: \(f(3) = 1(3)^2 - 6(3) + 10 = 9 - 18 + 10 = 1\) So, the vertex of the parabola is \(V(3,1)\).

3. Find the x-intercepts

The x-intercepts are the points where the parabola meets the x-axis, i.e., where \(f(x) = 0\). To find these, we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plug the coefficients into the formula: \(x=\frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)}\) Calculate the discriminant: \(D=(-6)^2 - 4(1)(10) = 36 - 40 = -4\) Since the discriminant is negative, the equation \(f(x) = 0\) has no real solutions. This means that the parabola doesn't intersect the x-axis.

4. Determine the regions where the inequality is true

We know that the vertex of the parabola is at point \(V(3,1)\), and the parabola doesn't intersect the x-axis. As the coefficient of the \(x^2\) term is positive, our parabola opens upwards. Therefore, the entire parabola lies above the x-axis, and the inequality \(f(x) > 0\) is true for all real values of x. The solution set for the inequality \(x^2 - 6x + 10 > 0\) is the entire real number line, which can be written using interval notation as \((-\infty, +\infty)\).

Key concepts

Quadratic Inequalities

When you come across a quadratic inequality like \(x^2 - 6x + 10 > 0\), you're looking for values of \(x\) where the quadratic expression is positive. A key point here is that a quadratic inequality doesn't only focus on finding specific points, like the solutions of an equation, but rather it's about determining entire ranges or intervals where the inequality holds true.
To visualize this better, imagine plotting the related quadratic equation \(y = x^2 - 6x + 10\) and looking for regions above the x-axis where \(y\) is positive. The graph of this function forms a parabola, and understanding its shape and position can tell us much about the inequality's solutions. Unlike equations, where we find where a graph intersects the axis, with inequalities we're interested in where the graph lies entirely above or below that axis.

Vertex of a Parabola

In a quadratic function given by \(y = ax^2 + bx + c\), the vertex represents the highest or lowest point of the parabola, depending on whether it opens upwards or downwards. To find this crucial point, use the formula \(V(\frac{-b}{2a}, f(\frac{-b}{2a}))\).
For the equation \(y = x^2 - 6x + 10\), we calculated the vertex as \(V(3,1)\). This information is not just geometrically significant; it also reveals the direction in which the parabola opens and helps us identify the range of the solutions for our inequality. The vertex, being the most extreme point on the parabola along the vertical axis, is either the maximum or minimum value of the quadratic function, essential for sketching the graph and understanding the inequality.

X-intercepts

The x-intercepts of a parabola are points where the graph crosses the x-axis, effectively where \(f(x) = 0\). These points are important as they can divide a number line into regions where an inequality may change from true to false or vice versa.
In the given quadratic inequality, we attempted to find x-intercepts using the quadratic formula but discovered there were none since the discriminant was negative. This tells us that the parabola does not cross the x-axis at any point. For inequalities, knowing whether or not the graph crosses the x-axis helps determine the intervals to be included in the solution set.

Quadratic Formula

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is a reliable method to find the x-intercepts of a parabola. It uses the coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(ax^2 + bx + c = 0\).
Applied to our inequality, when we crunched the numbers, the discriminant (\(b^2 - 4ac\)) was negative, indicating there are no real solutions and hence no x-intercepts. This lack of real solutions is significant: in terms of the inequality \(x^2 - 6x + 10 > 0\), it confirms that our parabola either sits entirely above or below the x-axis. Combined with the knowledge that the vertex is above the axis and the parabola opens upwards, it's clear that the inequality is satisfied for all real numbers.