Problem 31
Question
Find all real solutions of the equation. $$ x^{2}-7 x+10=0 $$
Step-by-Step Solution
Verified Answer
The real solutions are \( x = 5 \) and \( x = 2 \).
1Step 1: Identify the Equation Type
The equation given is a quadratic equation of the form \( ax^2 + bx + c = 0 \). In our case, \( a = 1 \), \( b = -7 \), and \( c = 10 \).
2Step 2: Use the Quadratic Formula
To solve the quadratic equation \( x^2 - 7x + 10 = 0 \), we can apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
3Step 3: Calculate the Discriminant
The discriminant \( \Delta \) is calculated as \( b^2 - 4ac \). Substitute the values: \( \Delta = (-7)^2 - 4 \times 1 \times 10 = 49 - 40 = 9 \).
4Step 4: Compute the Square Root of the Discriminant
Since the discriminant \( \Delta = 9 \), compute its square root: \( \sqrt{9} = 3 \).
5Step 5: Substitute into the Quadratic Formula
Now that we have \( \Delta \), substitute it into the formula: \[x = \frac{-(-7) \pm 3}{2 \times 1} = \frac{7 \pm 3}{2}.\]
6Step 6: Solve for the Two Possible Solutions
The plus-minus operation gives two solutions:1. \( x_1 = \frac{7 + 3}{2} = \frac{10}{2} = 5 \).2. \( x_2 = \frac{7 - 3}{2} = \frac{4}{2} = 2 \).
Key Concepts
Quadratic FormulaDiscriminantReal SolutionsStep by Step Solutions
Quadratic Formula
A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). The quadratic formula is a tool that allows us to find the solutions for these types of equations. It is given by the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides a way to directly calculate the roots of the equation without necessarily factoring or graphing.
To use it effectively:
To use it effectively:
- First, identify the coefficients \( a \), \( b \), and \( c \) in the equation.
- Plug these values into the formula.
- Solve for \( x \) using basic arithmetic operations to find the solutions.
Discriminant
The discriminant is a key part of the quadratic formula. It is found under the square root in the formula: \( b^2 - 4ac \). This value helps determine the nature of the roots of the quadratic equation.
When calculating the discriminant:
When calculating the discriminant:
- If the discriminant is positive, there are two distinct real solutions.
- If it is zero, there is exactly one real solution, which means both roots are the same.
- If it is negative, there are no real solutions, only complex solutions.
Real Solutions
Real solutions are the roots of the quadratic equation that can be plotted on a real-number line. They are the solutions that don't involve imaginary numbers. In our example, because the discriminant is positive (\( \Delta = 9 \)), we have two distinct real solutions: \( x_1 = 5 \) and \( x_2 = 2 \).
These solutions indicate the points where the quadratic polynomial intersects the x-axis. A positive discriminant ensures real roots, making the solutions observable in real-world contexts, such as projectile paths or economics models.
These solutions indicate the points where the quadratic polynomial intersects the x-axis. A positive discriminant ensures real roots, making the solutions observable in real-world contexts, such as projectile paths or economics models.
Step by Step Solutions
Step by step solutions provide clarity on why and how each part of the problem is solved. This approach breaks down complex processes into more manageable parts.
For the quadratic equation \( x^2 - 7x + 10 = 0 \):
For the quadratic equation \( x^2 - 7x + 10 = 0 \):
- We identified it as a quadratic equation with \( a = 1 \), \( b = -7 \), and \( c = 10 \).
- Calculated the discriminant \( b^2 - 4ac \), which was 9, a positive number.
- Used the quadratic formula, substituting \( \, b = -7 \, \) and \( \, c = 10 \, \) to find the roots.
- Determined the solutions \( x_1 = 5 \) and \( x_2 = 2 \) by computing \( \frac{7 \pm 3}{2} \).
Other exercises in this chapter
Problem 31
\(23-48\) Solve the inequality. Express the answer using interval notation. $$ |x+5| \geq 2 $$
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Solve the linear inequality. Express the solution using interval notation and graph the solution set. $$ -2
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\(5-60\) Find all real solutions of the equation. $$ \sqrt{5-x}+1=x-2 $$
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The given equation is either linear or equivalent to a linear equation. Solve the equation. \(\frac{2}{t+6}=\frac{3}{t-1}\)
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