Problem 86
Question
Solve the quadratic equation by entering the quadratic formula in the home screen of your graphing utility. (See Technology Note on page 167.) $$0.62 t^{2}-1.29 t+1.5=0$$
Step-by-Step Solution
Verified Answer
The solutions \(t\) are obtained by simplifying the quadratic equation using the quadratic formula and substituting in the values of \(a\), \(b\), and \(c\). The solutions will be two real numbers.
1Step 1: Substitute into Quadratic Formula
Substitute \(a = 0.62\), \(b = -1.29\) and \(c = 1.5\) into the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The equation becomes: \(t = \frac{1.29 \pm \sqrt{(-1.29)^{2} - 4*0.62*1.5}}{2*0.62}\)
2Step 2: Calculate Discriminant and Simplify Square Root
Next, calculate the value under the square root, which is the discriminant. Discriminant = \(b^{2} - 4ac = (-1.29)^{2} - 4 * 0.62 * 1.5\) Then, simplify the square root and overall equation.
3Step 3: Calculate Roots
After simplifying the determinate and rest of the equation, two roots will be found. These are the solutions for \(t\).
Key Concepts
Quadratic FormulaDiscriminantGraphing Utility
Quadratic Formula
Quadratic equations are essential parts of algebra, forming the foundation for understanding more complex polynomial functions. The quadratic formula is your go-to tool for solving these equations, which take the standard form of \(ax^{2}+bx+c=0\).
Why is the quadratic formula so amazing? It's a versatile formula that finds solutions to quadratic equations by plugging in the coefficients \(a\), \(b\), and \(c\). The formula is given by:
The quadratic formula saves time and provides exact values, eliminating guesswork. It is particularly handy when factoring is challenging or impossible. All you need are the coefficients \(a\), \(b\), and \(c\), and you're good to go! If you're looking to solve a quadratic equation, remember to double-check your inputs and calculate carefully to reach the correct roots.
Why is the quadratic formula so amazing? It's a versatile formula that finds solutions to quadratic equations by plugging in the coefficients \(a\), \(b\), and \(c\). The formula is given by:
- \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The quadratic formula saves time and provides exact values, eliminating guesswork. It is particularly handy when factoring is challenging or impossible. All you need are the coefficients \(a\), \(b\), and \(c\), and you're good to go! If you're looking to solve a quadratic equation, remember to double-check your inputs and calculate carefully to reach the correct roots.
Discriminant
The discriminant is the part of the quadratic formula under the square root, represented by \(b^2 - 4ac\). This term is significant because it indicates the number and type of solutions a quadratic equation might have.
- When the discriminant is positive, you'll find two distinct real roots. This means the parabola crosses the x-axis at two points.
- If the discriminant is zero, there is exactly one real root. This means the graph of the quadratic equation touches or is tangent to the x-axis.
- When the discriminant is negative, the quadratic equation has no real roots but instead two complex conjugate roots, implying the parabola doesn't intersect the x-axis at all.
Graphing Utility
Graphing utilities are wonderful technological aids for visualizing complex mathematical concepts, like quadratic equations. A graphing utility allows you to visualize equations on a coordinate plane, which can deepen your understanding of their behavior.
When solving a quadratic equation using the quadratic formula, graphing utilities become particularly useful to check and confirm solutions visually. Once you input the quadratic equation, the graphing tool displays its parabolic curve.
Here’s how a graphing utility can assist you:
When solving a quadratic equation using the quadratic formula, graphing utilities become particularly useful to check and confirm solutions visually. Once you input the quadratic equation, the graphing tool displays its parabolic curve.
Here’s how a graphing utility can assist you:
- Identify the roots: By observing where the parabola crosses the x-axis, you verify the solutions obtained from the quadratic formula.
- Understand graph symmetry: Visualizing the vertex and symmetry offers better insight into the equation’s properties.
- Experiment with different forms: Change the equation parameters \(a\), \(b\), and \(c\) to see real-time changes in the graph's shape and position.
Other exercises in this chapter
Problem 86
Let \(f(x)=2 x+5\) and \(g(x)=f(x+2)-4 .\) Graph both functions on the same set of coordinate axes. Describe the transformation from \(f(x)\) to \(g(x) .\) What
View solution Problem 86
In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{-x+1}{2
View solution Problem 87
In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(3 x-1)^{2}$$
View solution Problem 88
In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(-2 x+5)^{2}$$
View solution