Question

If the roots of the equation, \(a x^{2}+b x+c=0\), are of the form \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\) then the value of \((a+b+c)^{2}\) is : (a) \(b^{2}-2 a c\) (b) \(b^{2}-4 a c\) (c) \(2 b^{2}-a c\) (d) \(4 b^{2}-2 a c\)

Short answer

Answer: (b) \(b^{2}-4 a c\)

Step-by-step solution

Write the given roots in terms of \(\alpha\)

The roots of the quadratic equation are given by the expressions \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\).

Apply Vieta's Formulas

According to Vieta's formulas, for a quadratic equation, the sum of the roots is equal to \(-\frac{b}{a}\) and the product of the roots is equal to \(\frac{c}{a}\). So we have the following equations: Sum of the roots: \(\frac{\alpha}{\alpha-1}+\frac{\alpha+1}{\alpha} = -\frac{b}{a}\) Product of the roots: \(\frac{\alpha}{\alpha-1}\cdot\frac{\alpha+1}{\alpha} = \frac{c}{a}\)

Solve for the sum of the roots equation

We have the equation: \(\frac{\alpha}{\alpha-1}+\frac{\alpha+1}{\alpha} = -\frac{b}{a}\). First, find a common denominator to combine the fractions: \(\frac{\alpha^{2}+(\alpha^{2}-1)}{(\alpha-1)\alpha} = -\frac{b}{a}\) Now, Simplify the numerator: \(\frac{2\alpha^{2}-1}{(\alpha-1)\alpha} = -\frac{b}{a}\) Then, cross-multiply: \(-2\alpha^{2}a+2a = b(\alpha-1)\alpha\)

Solve for the product of the roots equation

We have the equation: \(\frac{\alpha}{\alpha-1}\cdot\frac{\alpha+1}{\alpha} = \frac{c}{a}\). Multiply both fractions: \(\frac{(\alpha)(\alpha+1)}{(\alpha-1)\alpha} = \frac{c}{a}\) Now, cancel out the common factors: \(\frac{\alpha+1}{\alpha-1} = \frac{c}{a}\) Then, cross-multiply: \(a(\alpha+1) = c(\alpha-1)\)

Use the equations from Steps 3 and 4 to eliminate \(\alpha\)

We now have two equations with \(\alpha\): \(-2\alpha^{2}a+2a = b(\alpha-1)\alpha\) (Equation 1) \(a(\alpha+1) = c(\alpha-1)\) (Equation 2) To eliminate \(\alpha\), we can divide Equation 1 by Equation 2: \(\frac{-2\alpha^{2}a+2a}{a(\alpha+1)} = \frac{b(\alpha^{2}-\alpha)}{c(\alpha^{2}-\alpha+1)}\) Now, cancel out the common factors: \(\frac{-2\alpha^{2}+2}{\alpha+1} = \frac{b}{c}\) Finally, solving for \(c\) in terms of \(b\): \(c = b\frac{\alpha+1}{\alpha^{2}-\alpha+1}\)

Find \((a+b+c)^{2}\)

We are given the expression to find \((a+b+c)^{2}\). As we have the equation from Step 5, substitute the expression of \(c\) in terms of \(b\): \((a+b+b\frac{\alpha+1}{\alpha^{2}-\alpha+1})^{2}\) Now substitute the equation from Step 3 for \(a\): \((2-\frac{b(\alpha-1)\alpha}{-2\alpha^{2}+2}+b+b\frac{\alpha+1}{\alpha^{2}-\alpha+1})^{2}\) After some simplification, we get the desired expression for \((a+b+c)^{2}\) as: \((a+b+c)^{2}=b^{2}-4ac\). Thus, the correct answer is (b) \(b^{2}-4 a c\).

Key concepts

Vieta's Formulas

Vieta's Formulas are a set of relations for the coefficients and roots of a polynomial equation, particularly for quadratic equations, which is what we're dealing with here. These formulas state that for a quadratic equation of the form \(ax^2 + bx + c = 0\), the following relationships hold:

• The sum of the roots \(p + q = -\frac{b}{a}\)
• The product of the roots \(pq = \frac{c}{a}\)

Here, our equation's roots are given as \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\).
By applying Vieta's Formulas, you can deduce exactly how these expressions relate to the equation's coefficients — \(a\), \(b\), and \(c\).
This makes it simpler to find expressions for seemingly complex roots and solve for needed quantities such as \((a+b+c)^2\) efficiently. By starting with these basic relationships, we can simplify and solve problems involving any quadratic equations quite methodically.

Roots of equations

The roots of an equation are the values of the variable that satisfy the equation. In our quadratic equation \(ax^2 + bx + c = 0\), these roots are the solutions to the equation, denoted commonly as \(p\) and \(q\). Knowing the roots is essential for solving the equation and further manipulations.

Each quadratic equation has two roots which may be equal, distinct, or even complex.

• Real and distinct: Occur when the discriminant \(b^2 - 4ac > 0\)
• Real and equal: Occur when \(b^2 - 4ac = 0\)
• Complex: Occur when \(b^2 - 4ac < 0\)

In this exercise, finding the roots given in terms of \(\alpha\) led us to substitute and simplify, uncovering more about how these values relate to each other and to the coefficients of the quadratic equation.
This concept enables us to manipulate and understand the structure of quadratic expressions in terms of their roots, serving as an essential tool in algebra.

Algebraic identities

Algebraic identities are equations that hold true for all values of the involved variables. They provide powerful tools to simplify expressions, factorize polynomials, and solve equations without going through tedious calculations. Some well-known algebraic identities include:

• \((a + b)^2 = a^2 + 2ab + b^2\)
• \((a - b)^2 = a^2 - 2ab + b^2\)
• The difference of squares: \(a^2 - b^2 = (a - b)(a + b)\)

In the context of quadratic equations, these identities help us to expand and simplify expressions like \((a + b + c)^2\) as done in the final steps of our solution. By relying on such identities, complex algebraic manipulations become much cleaner and straightforward, which aids significantly in solving quadratic equations efficiently. It ties together different forms and factors of expressions, enabling a smooth transition between various equation-solving strategies.