Problem 41
Question
Solve each equation using the Quadratic Formula. $$ x^{2}-6 x+9=0 $$
Step-by-Step Solution
Verified Answer
The solution to the equation is x = 3.
1Step 1: Determine the coefficients
Identify the coefficients of the quadratic equation in the standard form: ax^2+bx+c=0. Here, a=1, b=-6, and c=9.
2Step 2: State the Quadratic Formula
The Quadratic Formula is used to find the roots of a quadratic equation and is given by: x = (-b ± sqrt(b^2-4ac))/(2a).
3Step 3: Substitute the coefficients into the Quadratic Formula
Substitute a=1, b=-6, and c=9 into the formula: x = (-(-6) ± sqrt((-6)^2-4(1)(9)))/(2(1)).
4Step 4: Simplify under the square root
Simplify the expression under the square root: (-6)^2-4(1)(9)=36-36=0.
5Step 5: Solve for x
Since the square root of 0 is 0, the expression simplifies to x = (6 ± 0) / 2. Therefore, x=6/2.
6Step 6: Find the final result
Divide 6 by 2 to get the only solution for x, which is x = 3.
Key Concepts
Solving Quadratic EquationsRoots of a Quadratic EquationQuadratic Equation Coefficients
Solving Quadratic Equations
When tackling quadratic equations, the objective is to determine the values of x that satisfy the equation. These values are otherwise known as the 'roots' or 'solutions' of the equation.
A quadratic equation generally takes the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), are coefficients and \( a \) is not equal to zero. To find these roots, the Quadratic Formula, which is a reliable and universally applicable method, can be applied. In essence, this formula allows for a procedure that can locate the roots regardless of whether they are real or complex numbers.
The Quadratic Formula is stated as \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). To solve for x using this formula, you would carry out these steps: identify the coefficients \( a \), \( b \), and \( c \); substitute those values into the formula; simplify and calculate. The exercise provided showcases these steps in action, leading to the conclusion that \( x = 3 \) is the solution to the quadratic equation \( x^2 - 6x + 9 = 0 \).
A quadratic equation generally takes the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), are coefficients and \( a \) is not equal to zero. To find these roots, the Quadratic Formula, which is a reliable and universally applicable method, can be applied. In essence, this formula allows for a procedure that can locate the roots regardless of whether they are real or complex numbers.
The Quadratic Formula is stated as \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). To solve for x using this formula, you would carry out these steps: identify the coefficients \( a \), \( b \), and \( c \); substitute those values into the formula; simplify and calculate. The exercise provided showcases these steps in action, leading to the conclusion that \( x = 3 \) is the solution to the quadratic equation \( x^2 - 6x + 9 = 0 \).
Roots of a Quadratic Equation
The roots of a quadratic equation are the x-values at which the graph of the equation intersects the x-axis. These roots can be real or complex and there can be one or two of them.
In scenarios where the discriminant—given by \( b^2 - 4ac \) in the quadratic formula—is positive, there will be two distinct real roots. When the discriminant equals zero, it results in one real root, known as a repeated or double root. If the discriminant is negative, there are no real roots; instead, there are two complex roots.
The exercise provided is an example of an equation with a discriminant of zero (since \( 36 - 36 = 0 \)), leading to a double root at \( x = 3 \). This is why after applying steps from the Quadratic Formula, it concludes with just one solution, which graphically represents the point where the parabola touches the x-axis at a single point.
In scenarios where the discriminant—given by \( b^2 - 4ac \) in the quadratic formula—is positive, there will be two distinct real roots. When the discriminant equals zero, it results in one real root, known as a repeated or double root. If the discriminant is negative, there are no real roots; instead, there are two complex roots.
The exercise provided is an example of an equation with a discriminant of zero (since \( 36 - 36 = 0 \)), leading to a double root at \( x = 3 \). This is why after applying steps from the Quadratic Formula, it concludes with just one solution, which graphically represents the point where the parabola touches the x-axis at a single point.
Quadratic Equation Coefficients
In the expression \( ax^2 + bx + c = 0 \), the numbers \( a \), \( b \) and \( c \) represent the coefficients of the quadratic equation. The coefficient \( a \) is the leading coefficient and it determines the width and direction (up or down) of the parabola. The coefficient \( b \) influences the position of the vertex along the x-axis, while \( c \) represents the constant term and shows where the parabola intersects the y-axis.
Understanding each coefficient's role impacts how we visualize the graph of the quadratic equation. For instance, changing \( a \) alters the parabola's steepness and orientation; modifying \( b \) shifts the vertex left or right; and adjusting \( c \) lifts or lowers the graph vertically. It is also worth noting that the sign of \( a \) affects whether the parabola opens upwards (\(+\)) or downwards (\(-\)). In our exercise, \( a = 1 \) shows that the parabola opens upwards and is relatively wide, with \( b = -6 \) and \( c = 9 \) dictating the position of the vertex and y-intercept respectively.
Understanding each coefficient's role impacts how we visualize the graph of the quadratic equation. For instance, changing \( a \) alters the parabola's steepness and orientation; modifying \( b \) shifts the vertex left or right; and adjusting \( c \) lifts or lowers the graph vertically. It is also worth noting that the sign of \( a \) affects whether the parabola opens upwards (\(+\)) or downwards (\(-\)). In our exercise, \( a = 1 \) shows that the parabola opens upwards and is relatively wide, with \( b = -6 \) and \( c = 9 \) dictating the position of the vertex and y-intercept respectively.
Other exercises in this chapter
Problem 41
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