Problem 38
Question
Find all real solutions of the equation. $$ 8 x^{2}-6 x-9=0 $$
Step-by-Step Solution
Verified Answer
The solutions are \( x = \frac{3}{2} \) and \( x = \frac{-3}{4} \).
1Step 1: Identify the Quadratic Equation
The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, \( a = 8 \), \( b = -6 \), and \( c = -9 \).
2Step 2: Calculate the Discriminant
The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is calculated as \( b^2 - 4ac \). Substitute the values to find \( \Delta = (-6)^2 - 4 \times 8 \times (-9) = 36 + 288 = 324 \).
3Step 3: Evaluate the Discriminant
Since the discriminant \( \Delta = 324 \) is positive, the equation has two distinct real solutions.
4Step 4: Use the Quadratic Formula
The solutions are given by the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). Substitute \( b = -6 \), \( \Delta = 324 \), and \( a = 8 \) into the formula: \( x = \frac{6 \pm \sqrt{324}}{16} \).
5Step 5: Calculate the Square Root
Compute \( \sqrt{324} = 18 \). So the solutions simplify to \( x = \frac{6 \pm 18}{16} \).
6Step 6: Find the Two Solutions
Substitute the values into the formula:- For the plus sign: \( x = \frac{6 + 18}{16} = \frac{24}{16} = \frac{3}{2} \).- For the minus sign: \( x = \frac{6 - 18}{16} = \frac{-12}{16} = \frac{-3}{4} \).
Key Concepts
Discriminant of a Quadratic EquationQuadratic FormulaReal Solutions of Quadratic Equations
Discriminant of a Quadratic Equation
When solving a quadratic equation, the discriminant plays a crucial role. The discriminant is the part of the quadratic equation under the square root in the quadratic formula and is calculated using the formula: \( \Delta = b^2 - 4ac \). This small piece of the puzzle tells us how many real solutions we can expect from our quadratic equation.
- If \( \Delta > 0 \), there are two distinct real solutions.
- If \( \Delta = 0 \), there is exactly one real solution, also known as a repeated or double root.
- If \( \Delta < 0 \), no real solutions exist; instead, you have two complex solutions.
Quadratic Formula
The quadratic formula is a powerful tool used to find solutions of quadratic equations. The formula is given by:\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] It takes into account the coefficients from the standard quadratic form \( ax^2 + bx + c = 0 \):
- \( a \), the coefficient of \( x^2 \)
- \( b \), the coefficient of \( x \)
- \( c \), the constant term
Real Solutions of Quadratic Equations
Real solutions of a quadratic equation refer to the x-values where the quadratic graph intersects the x-axis. They are called real because they represent actual points on a number line, as opposed to complex solutions, which involve imaginary numbers.
Once you've plugged values into the quadratic formula and calculated \( \Delta \), it's about performing some basic calculations:
Once you've plugged values into the quadratic formula and calculated \( \Delta \), it's about performing some basic calculations:
- Calculate \( x = \frac{6 + 18}{16} = \frac{24}{16} = \frac{3}{2} \)
- Calculate \( x = \frac{6 - 18}{16} = \frac{-12}{16} = \frac{-3}{4} \)
Other exercises in this chapter
Problem 38
Evaluate the expression and write the result in the form a bi. $$ \frac{5-i}{3+4 i} $$
View solution Problem 38
Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ x(2-3 x) \leq 0 $$
View solution Problem 38
\(5-60\) Find all real solutions of the equation. $$ \left(\frac{x}{x+2}\right)^{2}=\frac{4 x}{x+2}-4 $$
View solution Problem 38
The given equation is either linear or equivalent to a linear equation. Solve the equation. \(\frac{6}{x-3}=\frac{5}{x+4}\)
View solution