Problem 5
Question
If the roots of the equation \(x^{2}-5 x+16=\) 0 are \(\alpha\) and \(\beta\), and the roots of equation \(x^{2}+p x+q=0\) are \(\alpha^{2}+\beta^{2}, \alpha \beta / 2\), then: (a) \(p=1, q=-56\) (b) \(p=-1, q=-56\) (c) \(p=1, q=56\) (d) \(p=-1, q=56\)
Step-by-Step Solution
Verified Answer
The answer is (a) \(p=1, q=-56\).
1Step 1: Identify Given Equation and Roots
The given quadratic equation is \(x^2 - 5x + 16 = 0\) with roots \(\alpha\) and \(\beta\). Using Vieta's formulas, we know: \(\alpha + \beta = 5\) and \(\alpha \beta = 16\).
2Step 2: Calculate \(\alpha^2 + \beta^2\)
Recall that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\). Substitute the known values: \((5)^2 - 2(16) = 25 - 32 = -7\). So, \(\alpha^2 + \beta^2 = -7\).
3Step 3: Calculate \(\alpha \beta / 2\)
Given \(\alpha \beta = 16\), we find \(\alpha \beta / 2 = 16 / 2 = 8\). So, \(\alpha \beta / 2 = 8\).
4Step 4: Form New Quadratic Equation
With roots \(\alpha^2 + \beta^2 = -7\) and \(\alpha \beta / 2 = 8\), the new quadratic equation is \(x^2 - (\alpha^2 + \beta^2 + \alpha \beta / 2)x + (\alpha^2 + \beta^2)(\alpha \beta / 2) = 0\).
5Step 5: Calculate Coefficient \(p\)
Calculate \(p = \alpha^2 + \beta^2 + \alpha \beta / 2\) using known values: \(p = -7 + 8 = 1\).
6Step 6: Calculate Coefficient \(q\)
Calculate \(q = (\alpha^2 + \beta^2)(\alpha \beta / 2) = (-7)(8) = -56\).
7Step 7: Match Values to Given Options
We found \(p = 1\) and \(q = -56\), which corresponds to option (a).
Key Concepts
Vieta's FormulasRoots of EquationsPolynomial Coefficients
Vieta's Formulas
In solving quadratic equations like the one given, Vieta's Formulas are a powerful tool. They help us establish connections between the coefficients of a polynomial and its roots. For a quadratic equation of the form \[ ax^2 + bx + c = 0 \]the roots, let's call them \( \alpha \) and \( \beta \), will satisfy:
- The sum \( \alpha + \beta = -\frac{b}{a} \)
- The product \( \alpha \beta = \frac{c}{a} \)
- \( \alpha + \beta = 5 \)
- \( \alpha \beta = 16 \)
Roots of Equations
Roots of an equation are the solution values for which the equation equals zero. For instance, in the polynomial given by \( x^2 - 5x + 16 = 0 \), the roots \( \alpha \) and \( \beta \) are those special numbers satisfying this expression. These roots can often be real or complex, and their nature affects the polynomial's behavior greatly.
When the given equation recognizes \( \alpha^2 + \beta^2 \) and \( \frac{\alpha \beta}{2} \) as its roots, it prompts the creation of a new quadratic equation. Solving this equation allows you to see how roots of one polynomial connect with those of another, all through a play of numbers and relationships. Don't forget:
When the given equation recognizes \( \alpha^2 + \beta^2 \) and \( \frac{\alpha \beta}{2} \) as its roots, it prompts the creation of a new quadratic equation. Solving this equation allows you to see how roots of one polynomial connect with those of another, all through a play of numbers and relationships. Don't forget:
- Real roots are those found on the x-axis of a graph.
- Complex roots come in pairs and appear as conjugates.
Polynomial Coefficients
Polynomial coefficients like \( a, b, \) and \( c \) in the equation \( ax^2 + bx + c \) not only set the shape of the graph but also play a crucial role in determining the polynomial's roots.
In the transitioning of roots and coefficients as seen in the problem's solution, substituting known values into relation formulas is crucial. Consider
In the transitioning of roots and coefficients as seen in the problem's solution, substituting known values into relation formulas is crucial. Consider
- The coefficient \( p \) is calculated by rearranging and using known root values: \( p = \alpha^2 + \beta^2 + \frac{\alpha \beta}{2} \).
- \( q \) is obtained by multiplying the expressions from calculated roots: \( q = (\alpha^2 + \beta^2) \times \frac{\alpha \beta}{2} \).
Other exercises in this chapter
Problem 5
Sum of roots of the quadratic equation is 2 and sum of cube of roots is 98 . Find the equation of roots.
View solution Problem 5
If one root of the equation \(5 x^{2}+13 x+k=0\) is reciprocal of other, then find the value of \(k\).
View solution Problem 6
If roots of equation \(2 x^{2}+3(k-2) x+4-k\) \(=15 x\) are same but of opposite sign, then fi nd the value of \(k\).
View solution Problem 7
If 8 and 2 are roots of the quadratic equation \(x^{2}+\alpha x+\beta=0\) and 3,3 are roots of the quadratic equation \(x^{2}+\alpha x+b=0 .\) Then, find the ro
View solution