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JEE Main 2019
Quadratic Equations
Quadratic Equation and Inequalities
Easy

Question

The value of aa for which the sum of the squares of the roots of the equation x2(a2)xa1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0 assume the least value is :

Options

Solution

Key Concepts and Formulas

  • Vieta's Formulas: For a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0 with roots α\alpha and β\beta, the sum of the roots is α+β=BA\alpha + \beta = -\frac{B}{A} and the product of the roots is αβ=CA\alpha \beta = \frac{C}{A}.
  • Sum of Squares of Roots: α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta
  • Vertex of a Parabola: The vertex of the parabola f(x)=ax2+bx+cf(x) = ax^2 + bx + c occurs at x=b2ax = -\frac{b}{2a}. If a>0a > 0, the vertex is a minimum.

Step-by-Step Solution

Step 1: Identify Coefficients and Apply Vieta's Formulas

The given quadratic equation is x2(a2)xa1=0x^2 - (a - 2)x - a - 1 = 0. We identify the coefficients as:

  • A=1A = 1
  • B=(a2)B = -(a - 2)
  • C=(a+1)C = -(a + 1)

Using Vieta's formulas, we find the sum and product of the roots α\alpha and β\beta:

  • Sum of roots: α+β=BA=(a2)1=a2\alpha + \beta = -\frac{B}{A} = -\frac{-(a - 2)}{1} = a - 2
  • Product of roots: αβ=CA=(a+1)1=(a+1)\alpha \beta = \frac{C}{A} = \frac{-(a + 1)}{1} = -(a + 1)

These expressions relate the sum and product of the roots to the parameter 'aa'.

Step 2: Express the Sum of Squares of Roots in Terms of 'aa'

We want to minimize α2+β2\alpha^2 + \beta^2. Using the identity α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta, we substitute the expressions for α+β\alpha + \beta and αβ\alpha \beta from Step 1: α2+β2=(a2)22((a+1))\alpha^2 + \beta^2 = (a - 2)^2 - 2(-(a + 1)) α2+β2=(a24a+4)+2(a+1)\alpha^2 + \beta^2 = (a^2 - 4a + 4) + 2(a + 1) α2+β2=a24a+4+2a+2\alpha^2 + \beta^2 = a^2 - 4a + 4 + 2a + 2 α2+β2=a22a+6\alpha^2 + \beta^2 = a^2 - 2a + 6 Let S(a)=a22a+6S(a) = a^2 - 2a + 6. This is the expression we want to minimize.

Step 3: Find the Value of 'aa' that Minimizes S(a)S(a)

S(a)=a22a+6S(a) = a^2 - 2a + 6 is a quadratic function. Since the coefficient of a2a^2 is positive, it has a minimum value. We can find the value of 'aa' that minimizes S(a)S(a) using the vertex formula: a=b2A=22(1)=22=1a = -\frac{b}{2A} = -\frac{-2}{2(1)} = \frac{2}{2} = 1 Alternatively, we can complete the square: S(a)=a22a+1+5=(a1)2+5S(a) = a^2 - 2a + 1 + 5 = (a - 1)^2 + 5 The minimum value of S(a)S(a) occurs when (a1)2=0(a - 1)^2 = 0, which means a=1a = 1.

Therefore, the sum of the squares of the roots is minimized when a=1a = 1.

Common Mistakes & Tips

  • Be careful with signs when applying Vieta's formulas and expanding expressions.
  • Remember to find the value of 'aa' that minimizes the sum of squares, not the minimum value itself.
  • Double-check algebraic manipulations to avoid errors.

Summary By applying Vieta's formulas, we expressed the sum of the squares of the roots as a quadratic function of 'aa'. Minimizing this quadratic function using the vertex formula, we found that the sum of the squares of the roots is minimized when a=1a = 1.

Final Answer The final answer is \boxed{1}, which corresponds to option (A).

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