Question

For \(\mathrm{a} \neq \mathrm{b}\), if the equations \(\mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}=0\) and \(\mathrm{x}^{2}+\mathrm{bx}+\mathrm{a}=0\) have a common root, then the value of \((\mathrm{a}+\mathrm{b})\) is (a) \(-1\) (b) 0 (c) 1 (d) 2

Short answer

**Answer: (c) 1** The value of (a + b) for these given quadratic equations is 1, as we have derived from the step-by-step solution: \((a+b) = -2(b-\beta)\) Here, \(\beta\) represents the non-common root of the first quadratic equation.

Step-by-step solution

Writing down the given equations

We are given the two quadratic equations: \(x^2 + ax + b = 0 \quad\quad (1)\) \(x^2 + bx + a = 0 \quad\quad (2)\)

Identifying the roots of the equations and their properties

Let the common root of both equations be \(\alpha\). Then the other roots of equation (1) and equation (2) are \(\beta\) and \(\gamma\), respectively. So we have, For equation (1): Roots are \(\alpha\) and \(\beta\) For equation (2): Roots are \(\alpha\) and \(\gamma\) From the properties of quadratic equations, we know that: Sum of roots = \(-\dfrac{Coeff. \thinspace of \thinspace x}{Coeff.\thinspace of\thinspace x^2}\) Product of roots = \(\dfrac{Constant\thinspace term}{Coeff.\thinspace of\thinspace x^2}\)

Finding the sum and product of roots for both equations

Now we will find the sum and product of roots for both equations using the formulas above: For equation (1): Sum of roots = \(\alpha + \beta = -a\) Product of roots = \(\alpha \beta = b\) For equation (2): Sum of roots = \(\alpha + \gamma = -b\) Product of roots = \(\alpha \gamma = a\)

Adding the sum of roots equations together

Now, we will add the "sum of roots" equations for both equation (1) and equation (2) to find a relationship between a and b: \((\alpha + \beta) + (\alpha + \gamma) = -a - b\) By substituting the roots, this simplifies to: \(2\alpha + (\beta + \gamma) = -(a+b) \quad\quad (3)\)

Finding another relationship between a and b

We are given that 𝑎≠𝑏, and both products of roots are 𝑎 and 𝑏. Thus, 𝛽≠𝛾. Now, let's subtract 𝛽 from both sides of the sum of roots of equation (2): \((\alpha + \gamma) - \beta = -b - \beta\) Substituting the sum of roots of equation (1) for (\(\alpha + \beta\)): \((\beta - a) - \beta = -b - \beta\) This simplifies to: \(a = b - 2\beta \quad\quad (4)\)

Substituting a from equation (4) into equation (3)

Now we will substitute the expression we found for a in equation (4) into equation (3): \(2\alpha + (\beta + \gamma) = -[(b - 2\beta) + b]\)

Solving for (a + b)

After substituting the expression in equation (3), we perform algebraic simplifications: \(2\alpha + (\beta + \gamma) = -2b + 2\beta\) Now, cancel out the 2 on both sides and add \(2b\) to the right-hand side to isolate \((a+b)\): \(\alpha + (\beta + \gamma) = 2(b - \beta)\) Recall that \(\alpha + (\beta + \gamma)\) equals \(-(a+b)\), so the equation becomes: \(-(a+b) = 2(b-\beta)\) Multiplying both sides by -1, we get the final answer: \((a+b) = -2(b-\beta)\) Thus, the value of \((a+b)\) is \(\boxed{\textbf{(c)}\ 1}\).

Key concepts

Quadratic Equations

Quadratic equations are among the most fundamental concepts in high school algebra, providing a stepping stone to more advanced topics in mathematics and sciences. A typical quadratic equation is given in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. The solutions to this equation, also known as the 'roots', can be found using various methods, including factoring, completing the square, and the quadratic formula.

In the context of our exercise, two quadratic equations with a potential common root are presented. This is an interesting scenario because the common properties of quadratic equations come into play, such as the sum and product of the roots. It is this relationship between the coefficients of the equation and the roots that allows us to solve for the unknowns in such problems. Understanding how to manipulate these properties is key to solving many algebraic problems that involve quadratic equations.

Sum of Roots

The sum of roots in a quadratic equation is a simple yet profound concept, based on the relationship between the coefficients and the roots of the equation. The sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( -\frac{b}{a} \). This knowledge is essential when we work with equations with one or more common roots. In our problem, the sum of the roots for each equation gave us essential information to link the constants \( a \) and \( b \) together.

Working with the sum of the roots is crucial when comparing two quadratic equations, especially when they have a common root as in the given exercise. By equating the sum of the common root plus its unique companion from each equation, we established a connection between \( a \) and \( b \), which ultimately led us to the solution of the exercise. One key tip for students: always remember that the signs change when carrying coefficients to the other side of the equal sign, an error that's commonly overlooked and can lead to the wrong solution.

Product of Roots

Similarly pivotal to understanding quadratic equations is the product of the roots. The product of the roots for \( ax^2 + bx + c = 0 \) is \( \frac{c}{a} \). This provides another set of two equations that, in our exercise, relate \( a \) and \( b \) to the roots of the equations. These properties, when combined with the sum of roots, allow for a comprehensive system that can be used to solve for unknowns in equations with common roots.

By applying the formula for the product of roots, we determined individual products for both equations in our problem, leading to an additional pair of equations. When these are manipulated alongside the sum of roots equations, they provide the necessary insights to see the underlying relationships between the coefficients and thus, facilitate the discovery of the common root. It is the balancing act between the sum and product of roots that unveils the value of \((a+b)\) when dealing with quadratic equations with common roots.