Problem 22
Question
If one root of the equation \(5 x^{2}+13 x+k=0\) is reciprocal of other, then find the value of \(k\).
Step-by-Step Solution
Verified Answer
The value of k for which one root of the equation \(5x^2+13x+k=0\) is the reciprocal of the other is k = 5.
1Step 1: Express the roots as reciprocals
Let α and β be the roots of the quadratic equation. Since one root is the reciprocal of the other, we can express them as α and 1/α.
2Step 2: Use Vieta's formulas to relate the roots to the coefficients
According to Vieta's formulas, for a quadratic equation of the form ax^2 + bx + c = 0:
Sum of roots = -b/a, and;
Product of roots = c/a.
In our case, a = 5, b = 13, and c = k. Therefore, we can write:
Sum of roots: α + 1/α = -13/5, and;
Product of roots: α * (1/α) = k/5.
3Step 3: Solve for the sum of the roots
We are given the equation, α + 1/α = -13/5. To solve this equation, we can multiply both sides by α:
α^2 + 1 = -13α/5
Now multiply both sides by 5 to remove the fraction:
5α^2 + 5 = -13α
Move all terms to the left side of the equation:
5α^2 + 13α + 5 = 0
4Step 4: Solve for the product of the roots
Since the roots are reciprocals, their product is 1 (α * 1/α = 1). So, we can say:
k/5 = 1
Now, solve for k:
k = 5
5Step 5: Conclusion
The value of k for which one root of the equation \(5x^2+13x+k=0\) is the reciprocal of the other is k = 5.
Key Concepts
Vieta's FormulasSum of RootsProduct of Roots
Vieta's Formulas
Vieta's formulas are a set of equations that relate the coefficients of a polynomial to sums and products of its roots. In the realm of quadratic equations, which take the general form of \(ax^2 + bx + c = 0\), Vieta's formulas state two key relationships: the sum of the roots \(\alpha + \beta\) is equal to \( -\frac{b}{a}\), and the product of the roots \(\alpha \beta\) is equal to \(\frac{c}{a}\).
These formulas are incredibly useful because they allow us to find relationships between the roots without actually solving the equation. For instance, if you're given the coefficients of a quadratic equation and you know the nature of its roots (such as one being the reciprocal of the other), you can immediately apply Vieta's formulas to set up equations for the sum and product of the roots. This shortcut bypasses the need for initial root calculation or completing the square.
These formulas are incredibly useful because they allow us to find relationships between the roots without actually solving the equation. For instance, if you're given the coefficients of a quadratic equation and you know the nature of its roots (such as one being the reciprocal of the other), you can immediately apply Vieta's formulas to set up equations for the sum and product of the roots. This shortcut bypasses the need for initial root calculation or completing the square.
Sum of Roots
In our exercise, Vieta's formula for the sum of roots \(\alpha + \beta = -\frac{b}{a}\) translates to an equation involving the roots of the given quadratic equation and its coefficients. The sum of roots reveals how the roots are connected.
In applying this to reciprocal roots, such as \(\alpha\) and \(\frac{1}{\alpha}\), the sum becomes \(\alpha + \frac{1}{\alpha}\). Our task, then, is to equate this expression to the negative coefficient of \(x\), divided by the coefficient of \(x^2\), and solve for the variable involved. This gives us a straightforward way to find one of the key characteristics of the equation, even before knowing the exact roots. It's also instrumental in identifying special relationships between the roots, such as one being the reciprocal of the other, as highlighted in our given problem.
In applying this to reciprocal roots, such as \(\alpha\) and \(\frac{1}{\alpha}\), the sum becomes \(\alpha + \frac{1}{\alpha}\). Our task, then, is to equate this expression to the negative coefficient of \(x\), divided by the coefficient of \(x^2\), and solve for the variable involved. This gives us a straightforward way to find one of the key characteristics of the equation, even before knowing the exact roots. It's also instrumental in identifying special relationships between the roots, such as one being the reciprocal of the other, as highlighted in our given problem.
Product of Roots
The product of the roots, according to Vieta's formulas, for a quadratic equation is \(\alpha \beta = \frac{c}{a}\). In the context of reciprocal roots, the product simplifies even more because if \(\alpha\) is a root, then its reciprocal is simply \(\frac{1}{\alpha}\), and their product is \(\alpha \times \frac{1}{\alpha} = 1\).
This simplification is crucial because it directly points to the value of the constant term \(c\) when \(a\) is known, without having to perform any complex calculations. In the exercise provided, by setting the product of roots to 1, we determine the unknown coefficient \(k\) of the equation. This aspect of quadratic equations is especially helpful when dealing with special cases, like reciprocal roots, making it easier to uncover specific properties of the equation.
This simplification is crucial because it directly points to the value of the constant term \(c\) when \(a\) is known, without having to perform any complex calculations. In the exercise provided, by setting the product of roots to 1, we determine the unknown coefficient \(k\) of the equation. This aspect of quadratic equations is especially helpful when dealing with special cases, like reciprocal roots, making it easier to uncover specific properties of the equation.
Other exercises in this chapter
Problem 20
If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+p x+1=0\) and \(\gamma, \delta\) the roots of \(x^{2}+q x+1=0\), show that \(q^{2}-p^{2}=(\alpha-\gamma)(\b
View solution Problem 21
If \(\sin \alpha\) and \(\cos \alpha\) are roots of the equation \(p x^{2}+q x+r=0\), then show that \(p^{2}-q^{2}+2 p r=0\).
View solution Problem 23
If the roots of \(p x^{2}+q x+2=0\) are reciprocal of each other, then find \(p\).
View solution Problem 24
Find the condition that the roots of the equation \(a x^{2}+b x+c=0\) be such that i. One root is \(n\) times the other. ii. One root is three times the other.
View solution