Problem 5
Question
If \(a, b\) are the roots of the equation \(x^{2}+x+1=0\), then find the value of \(a^{2}+b^{2}\).
Step-by-Step Solution
Verified Answer
The value of \(a^2 + b^2\) is \(-1\).
1Step 1: Identify the coefficients of the quadratic equation
We are given the quadratic equation, \(x^2 + x + 1 = 0\), where the coefficients are \(A = 1\), \(B = 1\), and \(C = 1\).
2Step 2: Apply Vieta's Formulas
Vieta's Formulas give us relationships between the roots and the coefficients of a quadratic equation. If a and b are the roots of the given equation, then we have:
1. Sum of roots: \(a + b = -\frac{B}{A}\)
2. Product of roots: \(ab = \frac{C}{A}\)
Now, we can substitute the coefficients we found in Step 1:
1. \(a + b = -\frac{1}{1} = -1\)
2. \(ab = \frac{1}{1} = 1\)
3Step 3: Find a^2 + b^2 using the relationships derived in Step 2
We already have the sum and product of roots, but we need the value of a^2 + b^2. We can derive this value using the relationships obtained in Step 2.
First, square the equation for the sum of roots:
\((a + b)^2 = (-1)^2\)
\(a^2 + 2ab + b^2 = 1\)
Next, substitute the value of the product of roots from Step 2 into the equation above:
\(a^2 + 2(1) + b^2 = 1\)
4Step 4: Solve for a^2 + b^2
Now we just need to isolate a^2 + b^2 in the equation above:
\(a^2 + b^2 = 1 - 2\)
\(a^2 + b^2 = -1\)
The value of \(a^2 + b^2\) is \(-1\).
Key Concepts
Quadratic EquationsSum and Product of RootsQuadratic Coefficients
Quadratic Equations
Quadratic equations are an essential part of algebra and they come in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients, and \(a \eq 0\). These equations are called 'quadratic' because \(quadra\) refers to the square in Latin, and the variable in the equation is squared (raised to the power of 2).
Quadratic equations can be solved by various methods including factoring, completing the square, using the quadratic formula, or graphing. However, an elegant way to understand the properties of a quadratic equation is through Vieta's Formulas, which provide a direct relationship between the roots of the equation and its coefficients. Mastery of these concepts is crucial to solve more complex problems involving quadratics.
Quadratic equations can be solved by various methods including factoring, completing the square, using the quadratic formula, or graphing. However, an elegant way to understand the properties of a quadratic equation is through Vieta's Formulas, which provide a direct relationship between the roots of the equation and its coefficients. Mastery of these concepts is crucial to solve more complex problems involving quadratics.
Sum and Product of Roots
The sum and product of the roots of a quadratic equation can be easily found using Vieta's Formulas, which are named after the Renaissance mathematician François Viète. For a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(\alpha\) and \(\beta\) (i.e., \(\alpha + \beta\)) is equal to \( -\frac{b}{a}\), and their product \(\alpha\beta\) equals \(\frac{c}{a}\).
These formulas come in handy, especially when you don't need to find the actual roots, but rather relationships between them, such as \(\alpha^2 + \beta^2\). In our exercise, knowing that \(a + b = -1\) and \(ab = 1\) allows us to find \(a^2 + b^2\) with simple algebraic manipulations. This not only simplifies the process but also gives insight into the symmetrical properties of the roots of quadratic equations.
These formulas come in handy, especially when you don't need to find the actual roots, but rather relationships between them, such as \(\alpha^2 + \beta^2\). In our exercise, knowing that \(a + b = -1\) and \(ab = 1\) allows us to find \(a^2 + b^2\) with simple algebraic manipulations. This not only simplifies the process but also gives insight into the symmetrical properties of the roots of quadratic equations.
Quadratic Coefficients
Quadratic coefficients, typically denoted as \(a\), \(b\), and \(c\) in the quadratic equation \(ax^2 + bx + c = 0\), serve as the backbone of the equation's properties and the solutions we derive. The coefficient \(a\) influences the concavity of the parabola represented by the quadratic equation, while \(b\) and \(c\) affect the location of its vertex and the y-intercept respectively.
Understanding the role of each coefficient deepens your comprehension of the equation's graph and provides the ability to predict the nature of its roots (real or complex), and their behavior. For instance, if the discriminant \(b^2 - 4ac\) is negative, the quadratic equation has no real roots and this reflects in the graph as the parabola does not cross the x-axis at all. In the context of our exercise, the coefficients are equal (\(a = b = c = 1\)), and we utilized them with Vieta's Formulas to deduce further relationships between the roots, bypassing the need for more complex solutions or graph-plots.
Understanding the role of each coefficient deepens your comprehension of the equation's graph and provides the ability to predict the nature of its roots (real or complex), and their behavior. For instance, if the discriminant \(b^2 - 4ac\) is negative, the quadratic equation has no real roots and this reflects in the graph as the parabola does not cross the x-axis at all. In the context of our exercise, the coefficients are equal (\(a = b = c = 1\)), and we utilized them with Vieta's Formulas to deduce further relationships between the roots, bypassing the need for more complex solutions or graph-plots.
Other exercises in this chapter
Problem 3
If the product of the roots of the equation \(m x^{2}+6 x+(2 m-1)=0\) is \(-1\), then find \(m\).
View solution Problem 4
If the sum of the roots of the equation \((a+1) x^{2}+(2 a+3) x+(3 a+4)=0\) is \(-1\), then find the product of the roots.
View solution Problem 6
If \(\alpha, \beta\) are the roots of the equation \(4 x^{2}+3 x+7=0\), then find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\).
View solution Problem 7
If \(\alpha\) and \(\beta\) are the roots of \(a x^{2}+b x+c=0\), find the values of the following:- i. \(\frac{1}{a \alpha+b}+\frac{1}{a \beta+b} .\left\\{\rig
View solution