Question

$x^2 + x + 1 + 2k(x^2 - x - 1) = 0$ is a perfect square for how many values of $k$

• A. 2
• B. 0
• C. 1
• D. 3

Answer 1

Answer: (A)

2

Answer 2

The problem asks for the number of values of $k$ for which the given equation $x^2 + x + 1 + 2k(x^2 - x - 1) = 0$ is a perfect square.

1. Rewrite the equation in standard quadratic form:

First, expand and rearrange the given equation into the standard quadratic form $Ax^2 + Bx + C = 0$:

$x^2 + x + 1 + 2kx^2 - 2kx - 2k = 0$

Group terms by powers of $x$:

$(1 + 2k)x^2 + (1 - 2k)x + (1 - 2k) = 0$

Comparing this with $Ax^2 + Bx + C = 0$, we identify the coefficients:

$A = 1 + 2k$ $B = 1 - 2k$ $C = 1 - 2k$

2. Apply the condition for a perfect square:

For a quadratic equation to be a perfect square (i.e., have equal roots), its discriminant ($D$) must be zero. The discriminant is given by the formula $D = B^2 - 4AC$.

Set the discriminant to zero:

$B^2 - 4AC = 0$

Substitute the expressions for $A$, $B$, and $C$:

$(1 - 2k)^2 - 4(1 + 2k)(1 - 2k) = 0$

3. Solve for k:

Notice that $(1 - 2k)$ is a common factor in both terms. Factor it out:

$(1 - 2k) [(1 - 2k) - 4(1 + 2k)] = 0$

This equation holds true if either of the factors is zero.

Case 1: The first factor is zero

$1 - 2k = 0$ $2k = 1$ $k = \frac{1}{2}$

Case 2: The second factor is zero

$(1 - 2k) - 4(1 + 2k) = 0$ $1 - 2k - 4 - 8k = 0$

Combine like terms:

$(1 - 4) + (-2k - 8k) = 0$ $-3 - 10k = 0$ $10k = -3$ $k = -\frac{3}{10}$

4. Count the number of values of k:

We found two distinct values for $k$: $k = \frac{1}{2}$ and $k = -\frac{3}{10}$.

Therefore, there are 2 values of $k$ for which the given equation is a perfect square.