 # Coefficients of a perfect square

(4x^4 – 4x^3 + Ax^2 – 8x + 16)  is a perfect square. What is the value of

the coefficient, A, of the x^2 term?

Only 161 out of 1,413 who participated in the Intermediate category of the 4th Philippine National Vedic Mathematics Olympiad were able to get the correct answer to this question.

Since it was given that this 4th degree polynomial is a perfect square, its square root will have the form (ax^2 + bx + c).

Using the First by the First, we have (ax^2)^2 = a^2x^4 = 4x^4. Therefore a = 2.

Next, we have, [2(2x^2)(bx)] = 4bx^3 = -4x^3. b, then must be -1.

The second to the last term, -8x is obtained from [2(bx)(c)] or 2(-x)(c) = -8x. c must be 4.

The square root must be (2x^2 – x + 4).

Ax^2 = [2(2x^2)4 + (-x)^2] = [16x^2 + x^2] = 17x^2, A = 17.

We can check our answer by using the sutra, the Product of the Sums is the Sum of the Product:

(2x^2 – x + 4)^2 = (4x^4 – 4x^3 + Ax^2 – 8x + 16).

Using the digit sums of the coefficients, we have

(2 – 1 + 4)^2 ?= (4 – 4 + 17 – 8 + 16)

or (5)^2 ?= 25; 25 = 25