Finding 'a' and 'b' for a Perfect Square Polynomial

We are given a polynomial \(x^4 + 2x^3 + ax^2 + bx + 9\) and told that it is a perfect square, where 'a' and 'b' are positive real numbers. A polynomial that is a perfect square can be written as the square of another polynomial.

Since the given polynomial is of degree 4, its square root must be a polynomial of degree 2. Let's assume the square root polynomial is of the form \(cx^2 + dx + e\).

So, we have: \[x^4 + 2x^3 + ax^2 + bx + 9 = (cx^2 + dx + e)^2\]

Let's expand the right side:

Now, we compare the coefficients of the expanded form with the coefficients of the given polynomial \(x^4 + 2x^3 + ax^2 + bx + 9\):

• Coefficient of \(x^4\): \(c^2 = 1\). This implies \(c = 1\) or \(c = -1\).
• Coefficient of \(x^3\): \(2cd = 2\).
• Coefficient of \(x^2\): \(d^2 + 2ce = a\).
• Coefficient of \(x\): \(2de = b\).
• Constant term: \(e^2 = 9\). This implies \(e = 3\) or \(e = -3\).

Step-by-Step Determination of 'a' and 'b'

Let's consider the possible values for c and e.

From \(c^2 = 1\), let's take \(c=1\).

From \(e^2 = 9\), we have two possibilities for e.

Case 1: \(c = 1\) and \(e = 3\)

• From \(2cd = 2\): \(2(1)d = 2 \implies 2d = 2 \implies d = 1\).
• From \(d^2 + 2ce = a\): \(a = (1)^2 + 2(1)(3) = 1 + 6 = 7\).
• From \(2de = b\): \(b = 2(1)(3) = 6\).

In this case, \(a = 7\) and \(b = 6\). The problem states that 'a' and 'b' must be positive real numbers. Since 7 and 6 are positive, this is a valid solution.

Case 2: \(c = 1\) and \(e = -3\)

• From \(2cd = 2\): \(2(1)d = 2 \implies 2d = 2 \implies d = 1\).
• From \(d^2 + 2ce = a\): \(a = (1)^2 + 2(1)(-3) = 1 - 6 = -5\).
• From \(2de = b\): \(b = 2(1)(-3) = -6\).

In this case, \(a = -5\) and \(b = -6\). These values are not positive, so this case is not valid according to the problem statement.

Considering \(c = -1\)

Let's check if taking \(c=-1\) yields any other valid solutions.

From \(2cd = 2\): \(2(-1)d = 2 \implies -2d = 2 \implies d = -1\). From \(e^2 = 9\), \(e=3\) or \(e=-3\).

Case 3: \(c = -1\) and \(e = 3\)

• From \(d^2 + 2ce = a\): \(a = (-1)^2 + 2(-1)(3) = 1 - 6 = -5\).
• From \(2de = b\): \(b = 2(-1)(3) = -6\).

Not positive, invalid.

Case 4: \(c = -1\) and \(e = -3\)

• From \(d^2 + 2ce = a\): \(a = (-1)^2 + 2(-1)(-3) = 1 + 6 = 7\).
• From \(2de = b\): \(b = 2(-1)(-3) = 6\).

In this case, \(a = 7\) and \(b = 6\). These are positive and match the values found in Case 1. The square root polynomial would be \(-x^2 - x - 3\), and \((-x^2 - x - 3)^2 = (x^2 + x + 3)^2\), which gives the same original polynomial.

Therefore, the only values for positive real numbers 'a' and 'b' that make the polynomial a perfect square are \(a = 7\) and \(b = 6\).

Verification

Let's check if \(x^4 + 2x^3 + 7x^2 + 6x + 9\) is indeed a perfect square. Based on our calculation, it should be \((x^2 + x + 3)^2\).

This matches the given polynomial with \(a=7\) and \(b=6\).

Revision Table: Perfect Square Polynomials

Additional Information: Properties of Polynomials

Understanding polynomial properties is key to solving such problems. Here are some related points:

• Degree of a product: The degree of the product of two polynomials is the sum of their degrees. If \(P(x)\) has degree m and \(Q(x)\) has degree n, then \(P(x)Q(x)\) has degree m+n. For a perfect square \( (Q(x))^2 \), if \(Q(x)\) has degree n, \((Q(x))^2\) has degree 2n.
• Roots of perfect squares: If a polynomial \(P(x)\) is a perfect square, its roots have an even multiplicity.
• Factoring: Recognizing that a polynomial is a perfect square is a type of polynomial factoring.
• Conditions for a perfect square: For a quadratic \(Ax^2+Bx+C\) to be a perfect square, the discriminant \(B^2-4AC\) must be zero. For higher-degree polynomials, the condition is that it can be written as \((Q(x))^2\).

By comparing coefficients, we systematically determine the unknown values 'a' and 'b' in the perfect square polynomial, ensuring they meet the criteria of being positive real numbers.