If the sum as well as the product of the roots of the equation px 2- 6x + q = 0 is 6, then what is (p + q) equal to?
7
Understanding the Quadratic Equation and its Roots
The given equation is \( px^2 - 6x + q = 0 \). This is a quadratic equation, which is generally represented in the standard form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are coefficients, and \( x \) is the variable.
For the given equation, by comparing it with the standard form, we can identify the coefficients:
- Coefficient of \( x^2 \), \( a = p \)
- Coefficient of \( x \), \( b = -6 \)
- Constant term, \( c = q \)
Properties of Roots: Sum and Product
For a quadratic equation \( ax^2 + bx + c = 0 \), if \( \alpha \) and \( \beta \) are the roots, then:
- The sum of the roots is given by \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots is given by \( \alpha \beta = \frac{c}{a} \)
Solving for Coefficients p and q
We are given that the sum of the roots of the equation \( px^2 - 6x + q = 0 \) is 6.
Using the formula for the sum of roots and the coefficients we identified:
\( \text{Sum of roots} = -\frac{b}{a} \)
\( 6 = -\frac{(-6)}{p} \)
\( 6 = \frac{6}{p} \)
To find the value of \( p \), we can multiply both sides by \( p \) and divide by 6:
\( 6p = 6 \)
\( p = \frac{6}{6} \)
\( p = 1 \)
We are also given that the product of the roots of the equation \( px^2 - 6x + q = 0 \) is 6.
Using the formula for the product of roots and the coefficients:
\( \text{Product of roots} = \frac{c}{a} \)
\( 6 = \frac{q}{p} \)
We have already found that \( p = 1 \). Substituting this value into the equation for the product of roots:
\( 6 = \frac{q}{1} \)
\( 6 = q \)
So, we have found that \( p = 1 \) and \( q = 6 \).
Calculating (p + q)
The question asks for the value of \( (p + q) \). Now that we know the values of \( p \) and \( q \), we can calculate their sum:
\( p + q = 1 + 6 \)
\( p + q = 7 \)
Therefore, the value of \( (p + q) \) is 7.
Revision Table: Quadratic Equation Roots
| Concept | Formula | Application to \( px^2 - 6x + q = 0 \) |
|---|---|---|
| Standard Form | \( ax^2 + bx + c = 0 \) | \( a=p, b=-6, c=q \) |
| Sum of Roots (\( \alpha + \beta \)) | \( -\frac{b}{a} \) | \( -\frac{(-6)}{p} = \frac{6}{p} \) |
| Product of Roots (\( \alpha \beta \)) | \( \frac{c}{a} \) | \( \frac{q}{p} \) |
Additional Information: Forming a Quadratic Equation
If the sum of the roots (\( S = \alpha + \beta \)) and the product of the roots (\( P = \alpha \beta \)) of a quadratic equation are known, the equation can be written as:
\( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \)
Or, \( x^2 - Sx + P = 0 \)
In this problem, the sum of roots is 6 and the product of roots is 6. So, the equation is \( x^2 - 6x + 6 = 0 \).
Comparing this with \( px^2 - 6x + q = 0 \):
- If we divide the given equation by \( p \) (assuming \( p \neq 0 \)), we get \( x^2 - \frac{6}{p}x + \frac{q}{p} = 0 \).
- Comparing \( x^2 - \frac{6}{p}x + \frac{q}{p} = 0 \) with \( x^2 - 6x + 6 = 0 \), we get:
- \( -\frac{6}{p} = -6 \implies \frac{6}{p} = 6 \implies p = 1 \)
- \( \frac{q}{p} = 6 \implies \frac{q}{1} = 6 \implies q = 6 \)
This confirms our values for \( p \) and \( q \), and the result \( p + q = 1 + 6 = 7 \).
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