Problem 143 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \((1-p)\) is a root of quadratic equation \(x^{2}+p x+\) \((1-p)=0\), then its roots are [2004] (A) 0,1 (B) \(-1,2\) (C) \(0,-1\) (D) \(-1,1\)

Step-by-Step Solution

Answer

The roots are 0 and -1.

Step 1: Substitute Root into the Equation

Given the quadratic equation \(x^2 + px + (1-p) = 0\), we know \((1-p)\) is a root. Substitute \(x = 1-p\) into the equation: \((1-p)^2 + p(1-p) + (1-p) = 0\).

Step 2: Expand and Simplify

Expand the terms: \((1-p)^2 = 1 - 2p + p^2\) and \(p(1-p) = p - p^2\). Substitute back into the equation: \[1 - 2p + p^2 + p - p^2 + 1 - p = 0\] which simplifies to \[2 - 2p = 0\].

Step 3: Solve for \(p\)

Solve \(2 - 2p = 0\) for \(p\) which results in \(2 = 2p\), thus \(p = 1\).

Step 4: Substitute \(p\) Back into the Original Quadratic

Substitute \(p = 1\) into the original equation: \(x^2 + 1x + (1-1) = 0\). This simplifies to \(x^2 + x = 0\).

Step 5: Factor the Quadratic Equation

Factor \(x^2 + x = 0\) as \(x(x + 1) = 0\). This gives roots \(x = 0\) and \(x = -1\).

Key Concepts

Roots of Quadratic Equation

Quadratic equations are frequently encountered in algebra, and understanding how to find their roots is a fundamental skill. The roots of a quadratic equation are the values of the variable that make the equation true or equal to zero. They are the solutions to the equation and are crucial because:

• They help in graphing the equation, particularly drawing the parabola.
• In real-world problems, they can represent important measurable quantities like time, distance, or profit.

For a general quadratic equation of the form \(ax^2 + bx + c = 0\), the roots can be found using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]However, in some cases, like the one in our exercise, a given root or particular conditions can help bypass this formula and lead to a more direct solution. Here, with one root \(1-p\), substituting it directly into the equation helps us find the specific value of \(p\) that ensures \(1-p\) is indeed a root.

Factoring Quadratics

When it comes to solving quadratic equations, factoring can be one of the simplest and most efficient methods. Factoring involves rewriting the quadratic equation as a product of two binomials, which can then be set to zero to find the roots.
The process works effectively when the quadratic can be expressed as \( (x - r)(x - s) = 0 \). The roots \( r \) and \( s \) are straightforward because setting each factor equal to zero will solve the equation.
In the exercise provided, after substituting and finding \(p = 1\), the quadratic equation simplifies to \(x^2 + x = 0\). This can be factored by taking \(x\) common, resulting in \(x(x + 1) = 0\). From here, the roots are easily determined as \(x = 0\) and \(x = -1\).

• Factoring is particularly useful when the equation is simple and integers are involved.
• This method can save time and avoid complex calculations.

Substituting Values in Equations

Substituting values into equations is a powerful technique to simplify expressions and find solutions. It involves replacing variables with known quantities or specific values to see how the equation behaves or to uncover insights.
In the provided exercise, substitution was key. We substituted \(1-p\) into the original quadratic equation \(x^2 + px + (1-p) = 0\). Through substitution, we expanded the equation and simplified to find a particular instance where the equation equals zero.
This led to obtaining the specific value \(p = 1\), ultimately simplifying the equation to be easily factored.

• Substitution can often reveal simplifications that aren't immediately obvious.
• It is a step-by-step approach ensuring each part of the equation is manageable.

Substituting values can turn complex processes into more straightforward calculations, making it an essential tool in problem-solving.