Let α and β be the roots of the equation \(\rm \frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\); a ≠ 0, b ≠ 0, x ≠ 0. Which one of the following is a quadratic equation whose roots are α2 and β2?
Understanding the Problem
The problem asks us to find a new quadratic equation. The roots of this new equation must be the squares of the roots (\(\alpha^2\) and \(\beta^2\)) of a given equation. The given equation is presented in a non-standard form, so our first step is to rearrange it into the standard quadratic form: \(\rm Ax^2 + Bx + C = 0\).
Step-by-Step Solution: Transforming the Given Equation
The given equation is:
\(\frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\)
We are given that \(a \ne 0\), \(b \ne 0\), and \(x \ne 0\). Let's rearrange this equation. A common strategy for equations with fractions is to combine terms or find a common denominator.
Let's move the \(\frac{1}{x}\) term to the left side:
\(\frac{1}{x+a+b} - \frac{1}{x} = \frac{1}{a}+\frac{1}{b}\)
Combine the terms on the left side by finding a common denominator, which is \(x(x+a+b)\):
\(\frac{x - (x+a+b)}{x(x+a+b)} = \frac{a+b}{ab}\)
Simplify the numerator on the left side:
\(\frac{x - x - a - b}{x(x+a+b)} = \frac{a+b}{ab}\)
\(\frac{-(a+b)}{x(x+a+b)} = \frac{a+b}{ab}\)
Assuming \(a+b \ne 0\) (if \(a+b=0\), the original equation simplifies to \(0=0\), which is not a quadratic equation with specific roots), we can cancel \(a+b\) from both sides:
\(\frac{-1}{x(x+a+b)} = \frac{1}{ab}\)
Now, cross-multiply:
\(-1 \times ab = 1 \times x(x+a+b)\)
\(-ab = x^2 + x(a+b)\)
Rearrange the terms to get the standard quadratic form:
\(x^2 + (a+b)x + ab = 0\)
This is the quadratic equation whose roots are \(\alpha\) and \(\beta\).
Using Vieta's Formulas for the Original Roots
For a quadratic equation \(\rm Ax^2 + Bx + C = 0\) with roots \(\alpha\) and \(\beta\), Vieta's formulas state:
- Sum of roots: \(\alpha + \beta = -\frac{B}{A}\)
- Product of roots: \(\alpha \beta = \frac{C}{A}\)
In our equation \(x^2 + (a+b)x + ab = 0\), we have \(A=1\), \(B=(a+b)\), and \(C=ab\). Therefore:
- \(\alpha + \beta = -\frac{a+b}{1} = -(a+b)\)
- \(\alpha \beta = \frac{ab}{1} = ab\)
Forming the New Quadratic Equation
The new quadratic equation has roots \(\alpha^2\) and \(\beta^2\). Let this new equation be \(x^2 - Sx + P = 0\), where S is the sum of the new roots and P is the product of the new roots.
- Sum of new roots: \(S = \alpha^2 + \beta^2\)
- Product of new roots: \(P = \alpha^2 \beta^2\)
We can express \(S\) and \(P\) in terms of the sum and product of the original roots:
\(S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(P = \alpha^2 \beta^2 = (\alpha \beta)^2\)
Now, substitute the values of \((\alpha + \beta)\) and \((\alpha \beta)\) we found using Vieta's formulas:
\(S = (-(a+b))^2 - 2(ab)\)
\(S = (a+b)^2 - 2ab\)
\(S = (a^2 + 2ab + b^2) - 2ab\)
\(S = a^2 + b^2\)
And for P:
\(P = (ab)^2 = a^2b^2\)
So, the new quadratic equation with roots \(\alpha^2\) and \(\beta^2\) is:
\(x^2 - Sx + P = 0\)
\(x^2 - (a^2 + b^2)x + a^2b^2 = 0\)
Comparing with Options
Let's compare our derived equation with the given options:
- Option 1: \(x^2 + (a^2 + b^2)x + a^2b^2 = 0\) (Incorrect sign for the middle term)
- Option 2: \(x^2 - (a^2 + b^2)x + a^2b^2 = 0\) (Matches our result)
- Option 3: \(x^2 - (a^2 + b^2)x - a^2b^2 = 0\) (Incorrect sign for the constant term)
- Option 4: \(x^2 + (a^2 + b^2)x - a^2b^2 = 0\) (Incorrect signs for both middle and constant terms)
Our derived equation \(x^2 - (a^2 + b^2)x + a^2b^2 = 0\) matches Option 2.
Revision Table: Key Steps and Formulas
| Step | Description | Formula/Result |
|---|---|---|
| 1 | Transform the given equation | \(x^2 + (a+b)x + ab = 0\) |
| 2 | Find sum of roots (\(\alpha+\beta\)) | \(-(a+b)\) |
| 3 | Find product of roots (\(\alpha\beta\)) | \(ab\) |
| 4 | Find sum of squared roots (\(\alpha^2+\beta^2\)) | \((\alpha+\beta)^2 - 2\alpha\beta = (a+b)^2 - 2ab = a^2+b^2\) |
| 5 | Find product of squared roots (\(\alpha^2\beta^2\)) | \((\alpha\beta)^2 = (ab)^2 = a^2b^2\) |
| 6 | Form new quadratic equation | \(x^2 - (\alpha^2+\beta^2)x + (\alpha^2\beta^2) = 0\) |
| 7 | Substitute values | \(x^2 - (a^2+b^2)x + a^2b^2 = 0\) |
Additional Information: Vieta's Formulas and Root Transformation
Vieta's formulas are a powerful tool that relates the coefficients of a polynomial equation to sums and products of its roots. For a quadratic equation \(Ax^2 + Bx + C = 0\), the sum of roots \(\alpha + \beta = -B/A\) and the product of roots \(\alpha\beta = C/A\).
These formulas are particularly useful when you need to find a new equation whose roots are some function of the original roots, without explicitly calculating the original roots themselves. In this problem, the new roots were \(\alpha^2\) and \(\beta^2\). We used the identities:
- \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
- \(\alpha^2 \beta^2 = (\alpha\beta)^2\)
These identities allowed us to express the sum and product of the new roots directly in terms of the sum and product of the original roots, which were easily found using Vieta's formulas on the transformed equation.
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