Q: The number of values of \( k \) for which \( p(x) = (16x^2 + 12x + 39) + k(9x^2-2x+11) \) is a perfect square is ?

\( \begin {aligned} a) &2 && b) 0 && c) 1 && d) \text{none of these} \end{aligned} \)

Ans: For \(p(x) \)to be a perfect square, its discriminant needs to be 0.

\(p(x) = (16+9k)x^2 + (12-2k)x + (39+11k)\) \(\therefore (6-k)^2 = (16+9k)(39+11k) \)

Solving the quadratic, we get 2 real roots, \(k \in \frac{-3}{2}, -4 \) So according to me, the answer should be 2.

But the answer is 0. How?

At \(k = -4, p(x) = (i\sqrt{20}x - i\sqrt{5})^2 \).

Also, we get real solutions for x. Then why no perfect square?

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TopNewest[Note: Harshit had a previous comment, which I was replying to and quoting from. This will make more sense had his comment not been deleted.]

The additional restriction is that \( g(x) \) must be a polynomial with integer coefficients, as you can see from my previous comment.

I don't see why \( 5x^2 + 10 x + 5 \) will be considered a perfect square, even though we can write \( 5 x^2 + 10 x + 5 = (\sqrt{5}x+\sqrt{5})^2\). The coefficients are not integers. If you consider \( 5x^2 + 10x + 5 \) is a perfect square, then it would suggest that for \(x=0\), you will consider that \( 5 \) is a perfect square. If this is the case, then the discussion is moot because you are working with very different definitions from everyone else.

You cannot say that you "took into account a square root complex coefficients as well, since its very rare to find perfect squares". In this case, you are no longer finding perfect squares, but a 'perfect Harshit square'. By changing the definitions of the question, you are no longer answering it properly. You also added irrational coefficients, hence, every non-negative real number is now a 'perfect Harshit square' since \( a = \sqrt{a} ^2 \). Adding complex coefficients means that every single complex number is now a 'perfect Harshit square'. – Calvin Lin Staff · 4 years, 8 months ago

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[Note: This is not a previous Brilliant problem.]

What is the definition of perfect square for polynomials?

If we look at the definition of perfect square for integers, even though \( 5 = \sqrt{5} ^2\) and \( -1 = i ^2\), we do not say that 5 or -1 is a perfect square. Instead, we require \( N = k^2\) is a perfect square if \(k\) is an integer.

In keeping with the ideas expressed in a perfect square of an integer, if we have \( f(x) = g(x)^2 \), we require \(g(x)\) to be a polynomial with integer coefficients. For example, \( 5 = \sqrt{5}^2\) or \( -x^2 = (ix)^2\) should not be considered a perfect square. [Though it seems from your example of \(p(x)\) that you will be willing to consider them as perfect squares.] Also \( x^3 = ( x^{\frac {3}{2} } )^2\) will not be considered a perfect square, since the \(g(x)\) is not a polynomial.

(Note: Some people consider it valid if the polynomial has rational coefficients, as opposed to integer coefficient. This depends on the context given. Most certainly, if the coefficients are irrational or imaginary, then it will not be a perfect square) – Calvin Lin Staff · 4 years, 8 months ago

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