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JEE Main 2020
Quadratic Equations
Quadratic Equation and Inequalities
Easy

Question

The minimum value of the sum of the squares of the roots of x2+(3a)x+1=2ax^{2}+(3-a) x+1=2 a 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, we have α+β=ba\alpha + \beta = -\frac{b}{a} and αβ=ca\alpha\beta = \frac{c}{a}.
  • Sum of Squares Identity: α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta
  • Completing the Square: A technique to rewrite a quadratic expression in the form a(xh)2+ka(x-h)^2 + k to easily find its minimum or maximum value.

Step-by-Step Solution

Step 1: Rewrite the equation in standard form

The given equation is x2+(3a)x+1=2ax^{2}+(3-a) x+1=2 a. To apply Vieta's formulas, we need to rewrite it in the standard quadratic form Ax2+Bx+C=0Ax^2 + Bx + C = 0. Subtracting 2a2a from both sides, we get: x2+(3a)x+(12a)=0x^{2}+(3-a) x+(1-2 a) = 0

Step 2: Identify coefficients and apply Vieta's formulas

Now, we can identify the coefficients:

  • A=1A = 1
  • B=(3a)B = (3-a)
  • C=(12a)C = (1-2a)

Let α\alpha and β\beta be the roots of the equation. Applying Vieta's formulas:

  • Sum of roots: α+β=BA=3a1=a3\alpha + \beta = -\frac{B}{A} = -\frac{3-a}{1} = a-3
  • Product of roots: αβ=CA=12a1=12a\alpha \beta = \frac{C}{A} = \frac{1-2a}{1} = 1-2a

Step 3: Express the sum of squares of roots in terms of 'a'

We want to find the minimum value of α2+β2\alpha^2 + \beta^2. Using the sum of squares identity: α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta Substitute the expressions for α+β\alpha + \beta and αβ\alpha\beta in terms of 'aa': α2+β2=(a3)22(12a)\alpha^2 + \beta^2 = (a-3)^2 - 2(1-2a)

Step 4: Simplify the expression and complete the square

Expand and simplify the expression: α2+β2=(a26a+9)2+4a\alpha^2 + \beta^2 = (a^2 - 6a + 9) - 2 + 4a α2+β2=a22a+7\alpha^2 + \beta^2 = a^2 - 2a + 7 Now, complete the square: α2+β2=(a22a+1)+6\alpha^2 + \beta^2 = (a^2 - 2a + 1) + 6 α2+β2=(a1)2+6\alpha^2 + \beta^2 = (a-1)^2 + 6

Step 5: Find the minimum value

The expression (a1)2(a-1)^2 is always non-negative. Its minimum value is 0, which occurs when a=1a = 1. Therefore, the minimum value of α2+β2\alpha^2 + \beta^2 is: (11)2+6=0+6=6(1-1)^2 + 6 = 0 + 6 = 6

Common Mistakes & Tips

  • Mistake: Forgetting to rearrange the equation into the standard form before applying Vieta's formulas.
  • Mistake: Sign errors when applying Vieta's formulas, especially for the sum of roots.
  • Tip: Completing the square is a powerful technique for finding the minimum or maximum value of a quadratic expression.

Summary

By rewriting the given quadratic equation in standard form and applying Vieta's formulas, we expressed the sum of the squares of the roots as a quadratic function of aa: a22a+7a^2 - 2a + 7. Completing the square, we found that this expression is equivalent to (a1)2+6(a-1)^2 + 6. The minimum value of this expression occurs when a=1a=1, and the minimum value is 6. This corresponds to option (C).

Final Answer

The final answer is \boxed{6}, which corresponds to option (C).

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