Question

For what values of k the expression $kx^2 + (k + 1)x + 3$ will be a perfect square of a linear polynomial.

Answer 1

k = 5 \pm 2\sqrt{6}

Answer 2

A quadratic expression of the form $Ax^2 + Bx + C$ is a perfect square of a linear polynomial if and only if its discriminant, $D = B^2 - 4AC$, is equal to zero.

Given the expression $kx^2 + (k + 1)x + 3$:
Here, we have $A = k$, $B = (k + 1)$, and $C = 3$.

For the expression to be a perfect square, its discriminant must be zero:
$D = (k + 1)^2 - 4(k)(3) = 0$

Expand and simplify the equation:
$(k^2 + 2k + 1) - 12k = 0$
$k^2 + 2k - 12k + 1 = 0$
$k^2 - 10k + 1 = 0$

This is a quadratic equation in terms of $k$. We can solve for $k$ using the quadratic formula:
$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For the equation $k^2 - 10k + 1 = 0$, we have $a=1$, $b=-10$, and $c=1$.

Substitute these values into the quadratic formula:
$k = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$
$k = \frac{10 \pm \sqrt{100 - 4}}{2}$
$k = \frac{10 \pm \sqrt{96}}{2}$

Simplify $\sqrt{96}$:
$\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}$

Substitute the simplified radical back into the expression for $k$:
$k = \frac{10 \pm 4\sqrt{6}}{2}$
Factor out 2 from the numerator:
$k = \frac{2(5 \pm 2\sqrt{6})}{2}$
$k = 5 \pm 2\sqrt{6}$

Thus, the values of $k$ for which the expression is a perfect square are $5 + 2\sqrt{6}$ and $5 - 2\sqrt{6}$.