For what values of k the expression kx²+(k+1)x+2 will be a perfect square of a linear polynomial.

To determine the values of k for which the expression kx² + (k + 1)x + 2 is a perfect square of a linear polynomial, we can follow these steps:

Step 1: Express the quadratic equation as a perfect square of a linear polynomial.
A perfect square of a linear polynomial takes the form (ax + b)² = a²x² + 2abx + b².

For our given expression, kx² + (k + 1)x + 2, we need to rewrite it in the form of (ax + b)².

Comparing the quadratic expression to the perfect square form, we have:
kx² + (k + 1)x + 2 = a²x² + 2abx + b²

Step 2: Equate the corresponding coefficients.
To find the values of a and b, we need to equate the corresponding coefficients of the two expressions.

a²x² = kx² ----> (1)
2abx = (k + 1)x ----> (2)
b² = 2 ----> (3)

Step 3: Solve the equations.
Now, let's solve the equations from Step 2.

Equation (1): a²x² = kx²
Since both sides of the equation are multiples of x², we can ignore x² for now and focus on the values of a and k.
a² = k ----> (4)

Equation (2): 2abx = (k + 1)x
Here, x is a common factor on both sides, so we can divide by x to simplify the equation.
2ab = k + 1 ----> (5)

Equation (3): b² = 2
Here, we need to solve for b.
Taking the square root of both sides, we get:
b = ±√2 (Note: We consider both positive and negative square roots) ----> (6)

Step 4: Find the values of k.
Now we have equations (4), (5), and (6) with three variables (a, b, k). We need to find the values of k that satisfy all these equations.

Substituting equation (6) into equation (5), we get:
2a(±√2) = k + 1
±2a√2 = k + 1
This equation tells us that k + 1 must be divisible by ±2√2.

Similarly, substituting equation (4) into equation (5), we get:
2a√k = k + 1
Here, k + 1 must be divisible by 2√k.

By combining the conditions, we find that k + 1 must be divisible by 2√k and ±2√2.

To simplify this, we can remove the square roots by squaring both sides of the equations.

From k + 1 being divisible by 2√k, we have:
(k + 1)² divisible by (2√k)²
(k + 1)² divisible by 4k

From k + 1 being divisible by ±2√2, we have:
(k + 1)² divisible by (±2√2)²
(k + 1)² divisible by 8

Now, the divisible conditions are:
(k + 1)² divisible by 4k and 8

To find the values of k that satisfy both conditions, we can try different values or use a numerical method.

After evaluation, we find that k = -7 and k = 0 are the solutions that satisfy both conditions.

So, for k = -7 and k = 0, the expression kx² + (k + 1)x + 2 is a perfect square of a linear polynomial.