Solve:
Consider the quadratic equation \( ax^2 + bx + c = 0 \), where a ≠ 0 , and its roots are \( \alpha \) and \( \beta \). To form a new quadratic equation whose roots are equal in magnitude but opposite in sign, the roots should be \( \alpha \) and \( -\alpha \).
The sum of the new roots: \( \alpha + (-\alpha) = 0 \).
The product of the new roots: \( \alpha \times (-\alpha) = -\alpha^2 \).
The required quadratic equation will be:
\(
x^2 - \)sum of roots \( x \) + product of roots \(= 0
\)
\(
x^2 - 0 \cdot x - \alpha^2 = 0 \implies x^2 - \alpha^2 = 0
\)
Solve:
Consider the quadratic equation \( 3x^2 - 8x + 5 = 0 \), whose roots are denoted by \( \alpha \) and \( beta \). We want to form an equation whose roots are equal in magnitude but opposite in sign, i.e., \( \alpha \) and \( -\alpha \).
The total of the roots from the initial equation:
\(
\alpha + \beta = \frac{8}{3}
\)
Product of roots:
\(
\alpha \beta = \frac{5}{3}
\)
For roots equal in magnitude but opposite in sign:
Sum = 0
Product = \( -\alpha^2 \).
To find \( \alpha^2 \), use the relation:
\(
(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2
\)
Since roots \( \beta = -\alpha \), sum is zero, so:
\(
\alpha + (-\alpha) = 0
\)
The quadratic with roots \( alpha \) and \( -alpha \) is:
\(
x^2 - ({sum}) x + ({product}) = x^2 - 0 + (-\alpha^2) = x^2 - \alpha^2 = 0
\)
Solve:
Suppose the roots of \( ax^2 + bx + c = 0 \) are \( \alpha \) and \( \beta \). Our objective is to derive a quadratic equation whose roots are each scaled by a constant factor \( k \), resulting in \( k\alpha \) and \( k\beta \).
The sum of new roots:
\(
k\alpha + k\beta = k(\alpha + \beta) = k \left(-\frac{b}{a}\right)
\)
The product of new roots:
\(
(k\alpha)(k\beta) = k^2 \alpha \beta = k^2 \frac{c}{a}
\)
Hence, the required quadratic equation is:
\(
x^2 - (k(\alpha + \beta)) x + k^2 \alpha \beta = 0
\)
Or substituting:
\(
x^2 - k \left(-\frac{b}{a}\right) x + k^2 \frac{c}{a} = 0
\)
Solve:
Given the quadratic equation:
\(
3x^2 - 5x - 1 = 0
\)
with roots \( \alpha \) and \( \beta \).
The roots of the new quadratic are \( 3\alpha \) and \( 3\beta \).
Using Vieta's formulas:
Sum of roots:
\(
\alpha + \beta = \frac{5}{3}
\)
Product of roots:
\(
\alpha \beta = \frac{-1}{3}
\)
Sum of new roots:
\(
3\alpha + 3\beta = 3(\alpha + \beta) = 3 \times \frac{5}{3} = 5
\)
Product of new roots:
\(
(3\alpha)(3\beta) = 9 \alpha \beta = 9 \times \frac{-1}{3} = -3
\)
The required quadratic equation with roots \( 3\alpha \) and \( 3\beta \) is:
\(
x^2 - (5)x - 3 = 0
\)
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