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Magazine Chapter 1: Problem 47
Solve the equation by any method. $$2 x^{2}=7 x+15$$
Short Answer
Expert verified
Answer: The solutions to the quadratic equation are $$x = 5$$ and $$x = -\frac{3}{2}$$.
Step by step solution
01
Set the equation equal to zero
Subtract 7x and 15 from both sides to rewrite the equation as $$2x^2 - 7x - 15 = 0$$.
02
Identify the coefficients
In the equation $$2x^2 - 7x - 15 = 0$$, we can identify the coefficients as $$a = 2$$, $$b = -7$$, and $$c = -15$$.
03
Apply the quadratic formula
The quadratic formula is given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Plug in the coefficients a, b, and c to find the solutions.
04
Substitute the coefficients in the quadratic formula
Using the coefficients $$a=2$$, $$b=-7$$, $$c=-15$$, the quadratic formula becomes $$x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-15)}}{2(2)}$$.
05
Simplify the formula
Simplify the quadratic formula to get $$x = \frac{7 \pm \sqrt{49 + 120}}{4}$$. Now, we can simplify further to get $$x = \frac{7 \pm \sqrt{169}}{4}$$.
06
Evaluate the square root
The square root of 169 is 13. Thus, the formula becomes $$x = \frac{7 \pm 13}{4}$$.
07
Solve for x
Now we have two possible solutions:
1. Adding 13: $$x = \frac{7 + 13}{4} = \frac{20}{4} = 5$$
2. Subtracting 13: $$x = \frac{7 - 13}{4} = \frac{-6}{4} = -\frac{3}{2}$$
So the solutions to the equation $$2x^2 = 7x + 15$$ are $$x = 5$$ and $$x = -\frac{3}{2}$$.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Quadratic Equations
A quadratic equation is an equation that can be rewritten in the form \( ax^2 + bx + c = 0 \). The challenge is to find the variable \( x \) that makes the equation true. Quadratic equations often appear in various mathematical and real-world problems, such as calculating areas and describing motion, so understanding how to solve them is crucial.
There are several methods to solve a quadratic equation:
There are several methods to solve a quadratic equation:
- Factoring: Express the equation as a product of its factors. This method is ideal when the equation is easily factorable.
- Completing the Square: Transform the equation into a perfect square. This method is usually used when the equation does not easily factor.
- Quadratic Formula: Apply a specific formula that provides a solution for any quadratic equation.
- Graphing: Plot the quadratic equation on a graph and identify the points where it crosses the x-axis.
Quadratic Formula
The quadratic formula is a powerful tool for finding the solutions of any quadratic equation. It is especially useful when factoring is impossible or difficult. The formula is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here is a breakdown of the components in the formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here is a breakdown of the components in the formula:
- \( a \), \( b \), and \( c \) are coefficients from the standard quadratic equation form \( ax^2 + bx + c = 0 \).
- The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It determines the nature and number of solutions.
- If the discriminant is positive, there are two distinct real roots.
- If the discriminant is zero, there is exactly one real root (or a repeated root).
- If the discriminant is negative, there are no real roots, but two complex roots.
- The symbol \( \pm \) indicates two solutions: one by adding and one by subtracting the square root.
Roots of Quadratic Equations
The solutions found from solving a quadratic equation are called its "roots." These roots are the values of \( x \) that satisfy the equation, meaning they make the equation equal zero. In our example, the roots are \( x = 5 \) and \( x = -\frac{3}{2} \).
Understanding the roots is essential:
Understanding the roots is essential:
- They represent the x-intercepts of the parabola formed by the quadratic equation when graphed.
- Roots can be real or complex, depending on the discriminant.
- If the equation factors perfectly, the roots can be found directly through factoring.
