Problem 40 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b\) are roots of the equation \(x^{2}+a b=(a+1) x\), then find the value of \(b\).

Step-by-Step Solution

Answer

The value of \(b\) is 1.

Step 1: Identify the coefficients of the quadratic equation

The given equation is \(x^{2} + a*b = (a + 1) * x\). However, the general form of a quadratic equation should look like: \(x^{2} + Bx + C = 0\). Rewrite the given equation so it's in the correct form: \(x^{2} - (a + 1) * x + a*b = 0\). Here, \(B = -(a + 1)\) and \(C = ab\).

Step 2: Apply Vieta's formulas

According to Vieta's formulas for roots of a quadratic equation, \(a + b\) is equal to \(-B\) and \(ab\) is equal to \(C\). So, \(a + b = a + 1\) and \(ab = ab\). In the first equation, if you subtract \(a\) from both sides, you'll get \(b = 1\). The second equation is trivially true.

Step 3: Write down the solution

The value of \(b\) that fulfills both conditions is \(b = 1\).

Key Concepts

Vieta's formulas

Vieta's formulas offer a clever shortcut for connecting the coefficients of a polynomial to its roots without requiring full solving of the equation. For a quadratic equation like:

• \(ax^2 + bx + c = 0\)

Vieta's formulas tell us:

• The sum of the roots \( (r_1 + r_2) \) is equal to \(-b/a\).
• The product of the roots \( (r_1 \times r_2) \) is equal to \(c/a\).

These formulas work because of the symmetry properties in the coefficients when the polynomial is expanded. They provide a quick way to review or predict root behaviors which can be particularly useful in complex problems. They allow us to determine basic relationships between the roots directly from the equation itself, without first finding the roots.

Roots of a polynomial

Roots of a polynomial are the values of the variable that make the polynomial equal to zero. In simpler terms, if you substitute these values into the polynomial equation, the result will be zero. In a quadratic equation, finding its roots is fundamental for understanding the behavior of the graph it represents.

• For instance, the roots of the equation \(x^2 - (a + 1)x + ab = 0\) are represented by \(a\) and \(b\).
• The roots are points where the graph touches or crosses the x-axis.

By finding these roots, you're locating points that help form the picture of the parabola. A parabola can open upwards or downwards depending on the sign of the leading coefficient. It's important to compare how these roots behave in different equations to get insight into possible real-world applications. The process of finding roots varies depending on the complexity of the polynomial but quadratic equations are among the simplest to work with.

Solving quadratic equations

Solving quadratic equations involves finding the values of the variable that satisfy the equation. A common quadratic equation looks like \(ax^2 + bx + c = 0\). Various methods can be used to solve these equations:

• Factoring: If the equation can be factored easily into binomials, the roots can be found by setting each factor equal to zero and solving for the variable.
• Quadratic Formula: When factoring is difficult, the quadratic formula is a savior. It is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) and always works for any quadratic equation.
• Completing the Square: This method turns the quadratic into a perfect square trinomial, which can then be solved easily.
• Graphical Method: Plotting the equation to find where it intersects the x-axis can give a visual representation of the roots.

Each method has its applications and advantages depending on the equation structure. The choice of method often depends on the intended use or specific conditions given in an exercise, such as symmetry, ease, or speed preference when calculating. Quadratics play an essential role in various aspects of mathematics and science, which makes understanding their solutions pivotal.