Minimum Value of Sum of Squares of Roots

CAT 2017 Slot 2 · QA · Medium · Quadratic Equations

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The minimum possible value of the sum of the squares of the roots of the equation x2+(a+3)x−(a+5)=0x^2 + (a + 3)x - (a + 5) = 0 is

  1. A.

    1

  2. B.

    2

  3. C.

    3

  4. D.

    4

Answer

C

Explanation

Let the roots be α\alpha and β\beta.

α+β=−(a+3)\alpha + \beta = -(a + 3) αβ=−(a+5)\alpha \beta = -(a + 5)

Sum of squares of roots S=α2+β2=(α+β)2−2αβS = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta: S=[−(a+3)]2−2[−(a+5)]=a2+6a+9+2a+10=a2+8a+19S = [-(a+3)]^2 - 2[-(a+5)] = a^2 + 6a + 9 + 2a + 10 = a^2 + 8a + 19

To find the minimum value of a2+8a+19a^2 + 8a + 19: S=(a+4)2+3S = (a + 4)^2 + 3

Since (a+4)2≥0(a+4)^2 \ge 0, the minimum value is 3 (occurring when a=−4a = -4).

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