Problem 17
Question
Solve the given quadratic equations using the quadratic formula. If there are no real roots, state this as the answer. Exercises \(3-6\) are the same as Exercises \(13-16\) of Section 7.2. $$30 y^{2}+23 y-40=0$$
Step-by-Step Solution
Verified Answer
The solutions are \( y = \frac{5}{6} \) and \( y = \frac{-8}{5} \).
1Step 1: Identify the coefficients
The given equation is \(30y^2 + 23y - 40 = 0\). The quadratic equation is in the form \(ay^2 + by + c = 0\). Here, identify the coefficients as \(a = 30\), \(b = 23\), and \(c = -40\).
2Step 2: Recall the quadratic formula
The quadratic formula to find the roots of the equation \(ax^2 + bx + c = 0\) is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We will use this formula to find the roots of the equation.
3Step 3: Calculate the discriminant
The discriminant \(\Delta\) is given by \(b^2 - 4ac\). Substitute the values to find \(b^2 - 4ac = 23^2 - 4 \cdot 30 \cdot (-40)\). Calculate this to find \(\Delta = 529 + 4800 = 5329\).
4Step 4: Determine the nature of the roots
Since the discriminant \(\Delta = 5329\) is greater than zero, the quadratic equation will have two distinct real roots.
5Step 5: Substitute in the quadratic formula
Use the quadratic formula: \( y = \frac{-23 \pm \sqrt{5329}}{60} \). Calculate \(\sqrt{5329}\) which is 73.
6Step 6: Calculate each root
There are two roots to find: 1. \( y_1 = \frac{-23 + 73}{60} = \frac{50}{60} = \frac{5}{6}\)2. \( y_2 = \frac{-23 - 73}{60} = \frac{-96}{60} = \frac{-8}{5}\)
Key Concepts
Quadratic FormulaDiscriminantReal RootsCoefficients
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It allows us to find the roots, or solutions, of these equations efficiently. The formula is given by:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Discriminant
The discriminant is a key part of the quadratic formula, nestled inside the square root. It is given by \(b^2 - 4ac\). The value of the discriminant helps determine the nature and number of roots an equation has:
- If \(\Delta > 0\), the equation has two distinct real roots.
- If \(\Delta = 0\), there is one real root (also called a repeated or double root).
- If \(\Delta < 0\), there are no real roots; the roots are complex or imaginary.
Real Roots
Real roots are the solutions to a quadratic equation that can be represented on the real number line. If a quadratic equation has real roots, it means that the graph of the equation intersects the x-axis at real number points.
- Real roots can be distinct, meaning the graph intersects the x-axis at two separate points.
- Or they can coincide when they are the same number, resulting in a graph that just touches the x-axis at one point.
Coefficients
In a quadratic equation like \(30y^2 + 23y - 40 = 0\), the terms \(a\), \(b\), and \(c\) are referred to as the coefficients. Each of these plays a crucial role in determining the shape and position of the quadratic equation's graph:
- \(a\) is the coefficient of the quadratic term \(y^2\). It influences the direction and width of the parabola. If \(a\) is positive, the parabola opens upwards. If negative, it opens downwards.
- \(b\) is the coefficient of the linear term \(y\), affecting the position of the vertex and axis of symmetry.
- \(c\) is the constant term, which moves the parabola up or down on the graph.
Other exercises in this chapter
Problem 16
Solve the given quadratic equations by factoring. $$x^{2}+x-6=0$$
View solution Problem 16
Solve the given quadratic equations by completing the square. $$t^{2}+5 t-6=0$$
View solution Problem 17
Use a calculator to solve the given equations. Round solutions to the nearest hundredth. If there are no real roots, state this. $$2 x^{2}-7=0$$
View solution Problem 17
Solve the given quadratic equations by factoring. $$R^{2}+12=7 R$$
View solution