Problem 79
Question
In Exercises \(75-82\), compute the discriminant. Then determine the number and type of solutions for the given equation. $$x^{2}-2 x+1=0$$
Step-by-Step Solution
Verified Answer
The discriminant of the equation \(x^{2}-2x+1=0\) is zero, indicating it has one real solution or repeated root.
1Step 1: Calculate the discriminant
First, calculate the discriminant using the formula \(D=b^{2}-4ac\). So we substitute \(a=1\), \(b=-2\), and \(c=1\) into the formula which gives us \(D=(-2)^{2}-4*1*1 = 4-4 = 0\)
2Step 2: Determine the number and type of solutions
Since the discriminant (D) is zero, that means the equation has one real solution or repeated root.
Key Concepts
Quadratic EquationsDiscriminant ComputationTypes of SolutionsAlgebraic Solutions
Quadratic Equations
Quadratic equations are polynomial equations of the second degree, which means they include a variable (\(x\)) raised to the power of two. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \) where \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \) and \( c \) is the constant term. Solutions to these equations are the values of \( x \) that make the equation true, and these values are known as the roots of the equation.
Discriminant Computation
The discriminant is a key component in determining the nature and number of solutions to a quadratic equation. It is computed using the formula \( D = b^2 - 4ac \) where \( a \) is the coefficient of the \( x^2 \) term, \( b \) is the coefficient of the \( x \) term, and \( c \) is the constant. The value of the discriminant tells us whether the quadratic equation has two distinct real solutions, one repeated real solution, or two complex solutions.
Types of Solutions
Based on the discriminant (\( D \) ), quadratic equations can have different types of solutions:
- If \( D > 0 \) the equation has two distinct real solutions.
- If \( D = 0 \) there is exactly one real solution, also known as a repeated or double root.
- If \( D < 0 \) then the equation has no real solutions but two distinct complex solutions.
Algebraic Solutions
Algebraic solutions refer to the actual values of \( x \) that satisfy the quadratic equation. These can be found by various methods, including factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \) where \( D \) is the discriminant. The method you choose may depend on the specific form of the quadratic equation and the value of the discriminant. For the equation \( x^2 - 2x + 1 = 0 \) with discriminant \( D = 0 \) we find that \( x = 1 \) is the singular real solution, obtained either by factoring to \( (x-1)^2 = 0 \) or using the quadratic formula.
Other exercises in this chapter
Problem 79
Solve absolute value inequality. \(3|x-1|+2 \geq 8\)
View solution Problem 79
combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, o
View solution Problem 79
Exercises \(78-80\) will help you prepare for the material covered in the next section. Factor: $$ x^{2}-6 x+9 $$
View solution Problem 80
Solve absolute value inequality. \(5|2 x+1|-3 \geq 9\)
View solution