Problem 101 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The set of values of \(a\) for which the equation \(\left(x^{2}+x\right.\) \(+2)^{2}-(a-3)\left(x^{2}+x+2\right)\left(x^{2}+x+1\right)+(a-4)\left(x^{2}+\right.\) \(x+1)^{2}=0\) has at least one real root is (A) \(\left(5, \frac{19}{3}\right)\) (B) \(\left[5, \frac{19}{3}\right]\) (C) \(\left[5, \frac{19}{3}\right)\) (D) \(\left(5, \frac{19}{3}\right]\)

Step-by-Step Solution

Answer

The set of values for \(a\) is (C) \( [5, \frac{19}{3}) \).

Step 1: Simplify the Original Equation

The given equation is \( \left(x^2 + x + 2\right)^2 - (a - 3)\left(x^2 + x + 2\right)\left(x^2 + x + 1\right) + (a - 4)\left(x^2 + x + 1\right)^2 = 0 \). Introduce a substitution \( y = x^2 + x + 1 \), making the expression \( \left(y+1\right)^2 - (a-3)(y+1)y + (a-4)y^2 = 0 \). Expand \( \left(y+1\right)^2 = y^2 + 2y + 1 \) and substitute back.

Step 2: Form the Quadratic Equation in terms of y

Substitute and rearrange: \( y^2 + 2y + 1 - (a - 3)(y^2 + y) + (a-4)y^2 = 0 \). Simplify the expression: \( y^2 + 2y + 1 - (a-3)y^2 - (a-3)y + (a-4)y^2 = 0 \). Combine terms: \((3-a)y^2 + (-a-1)y + 1 = 0 \). This is the quadratic equation in \( y \).

Step 3: Determine Conditions for Real Roots in y

To have at least one real root for \( y \), the discriminant \( b^2 - 4ac \) of the quadratic equation \( ay^2 + by + c = 0 \) must be non-negative. For \((3-a)y^2 + (-a-1)y + 1 = 0 \), \( a = 3-a \), \( b = -(a+1) \), and \( c = 1 \). The discriminant is \((-a-1)^2 - 4(3-a)(1) \geq 0 \).

Step 4: Solve the Discriminant Inequality

Calculate: \((-a-1)^2 = a^2 + 2a + 1 \) and \(4(3-a) = 12 - 4a \). The inequality becomes \( a^2 + 2a + 1 - (12 - 4a) \geq 0 \). Simplify: \( a^2 + 6a - 11 \geq 0 \).

Step 5: Factor and Solve Quadratic Inequality

Solve \( a^2 + 6a - 11 = 0 \) to find roots: Use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \equiv a = \frac{-6 \pm \sqrt{36 + 44}}{2} \equiv a = \frac{-6 \pm \sqrt{80}}{2} \). Simplify \( \sqrt{80} = 4\sqrt{5} \), so the roots are \( a = \frac{-6 \pm 4\sqrt{5}}{2} \equiv a = -3 \pm 2\sqrt{5} \).

Step 6: Determine the Interval for a

The critical values, \( a = -3 + 2\sqrt{5} \approx 5.472 \) and \( a = -3 - 2\sqrt{5} \), dictate the intervals where \( a^2 + 6a - 11 \geq 0 \). Consider the polynomial sign changes: \( a \geq 5.472 \) for real roots. Comparing to the answer options, this translates to the interval \( \left[5, \frac{19}{3} \right) \) for \( a \).

Key Concepts

Real Roots

A real root of a quadratic equation is a solution where the value of the variable results in a real number. In simple words, when we solve the quadratic equation, if we end up with real numbers rather than complex or imaginary numbers, we say the equation has real roots. Real numbers are any numbers that you are familiar with: fractions, whole numbers, decimals, and so forth, which include both positive and negative numbers. Quadratic equations can have:

• Two distinct real roots
• One real double root
• No real roots, only complex roots

For students, it is crucial to understand when a quadratic equation will have real roots. This typically depends on another important concept called the discriminant.

Discriminant

In a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant helps to determine the nature of the roots. The discriminant is expressed as \( b^2 - 4ac \). It is a powerful tool that tells us:

• If the discriminant \( > 0 \), there are two distinct real roots.
• If the discriminant \( = 0 \), there is exactly one real root, a double root.
• If the discriminant \( < 0 \), there are no real roots; instead, there are two complex roots.

In the context of our problem, we formed a quadratic in \( y \), which transformed into a discriminant inequality \( (-a-1)^2 - 4(3-a)(1) \geq 0 \). Solving this inequality helped us determine the range of values for which at least one real root exists.

Substitution Method

The substitution method is a useful technique to simplify complex algebraic equations by introducing a new variable. In this exercise, substitution was used to simplify the given equation by letting \( y = x^2 + x + 1 \). This substitution effectively reduced the complexity and helped transform the equation into a simpler form, which in this case became a quadratic in \( y \).Here's how substitution aids in problem-solving:

• Identifies a common part of the equation that can be replaced by a new variable, simplifying the equation.
• Transforms complex terms into manageable ones.
• Helps reveal the structure of the mathematical problem, making it easier to solve.

Once simplified, you can solve for the new variable and back-substitute to find the original variable's solutions. Thus, the substitution method is essential for solving certain quadratic equations efficiently.