For how many <strong>REAL</strong> values of k is 6kx^{2}+12kx-24x+16 a perfect square for <strong>EVERY</strong> integer x?

Explanation

The expression is 6kx^2 + (12k-24)x + 16. For it to be a perfect square for all x, its discriminant must be zero: (12k-24)^2 - 4(6k)(16) = 0. Simplifying gives 144(k-2)^2 - 384k = 0, which reduces to 3k^2 - 20k + 12 = 0. Solving yields k = 6 and k = 2/3. There are two real values.

Related questions on Algebra

• If p + q + r = 0, then what is z^{\frac{p^2}{qr}} \times z^{\frac{q^2}{rp}} \times z^{\frac{r^2}{pq}} equal to ?
• What is the value of k for which (k^2 - 5k + 4)x^2 + (k^2 - 3k - 4)x + (k^2 - 4k) = 0 is an identity ?
• If \frac{a^2}{b^2 + c^2} = \frac{b^2}{c^2 + a^2} = \frac{c^2}{a^2 + b^2}, then what is the value of a^4 + b^4 + c^4 equal to ?
• If \frac{1}{x} = \frac{1}{p} + \frac{1}{q}, then what is \frac{pq}{p^2 - q^2}\left(\frac{x + p}{x - p} - \frac{x + q}{x - q}\right) equa...
• If (x - 5) is the HCF of x^2 - x - p and x^2 - qx - 10, then what is the value of (p + q) ?