Understanding Quadratic Equations and Common Roots

The problem asks us to find the value of \(a + b\) given two quadratic equations that share a common root. The equations are:

1. \(x^2 + ax + b = 0\)
2. \(x^2 + bx + a = 0\)

We are also told that \(a\) is not equal to \(b\), which is an important condition.

Finding the Common Root

Let the common root be denoted by \(\alpha\). Since \(\alpha\) is a root of both equations, substituting \(\alpha\) for \(x\) in both equations must satisfy them:

• \(\alpha^2 + a\alpha + b = 0\) (Equation 1)
• \(\alpha^2 + b\alpha + a = 0\) (Equation 2)

To find the value of the common root \(\alpha\), we can eliminate the \(\alpha^2\) term by subtracting one equation from the other. Subtracting Equation 2 from Equation 1:

\((\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0\)

This simplifies to:

\(\alpha^2 + a\alpha + b - \alpha^2 - b\alpha - a = 0\)

\(a\alpha - b\alpha + b - a = 0\)

Solving for the Common Root Value

We can factor the terms involving \(\alpha\) and the constant terms separately:

\(\alpha(a - b) - (a - b) = 0\)

Now, we can factor out the term \((a - b)\):

\((a - b)(\alpha - 1) = 0\)

We are given that \(a \neq b\), which means \(a - b \neq 0\). For the product of two factors to be zero, at least one of the factors must be zero. Since \((a - b)\) is not zero, the other factor, \((\alpha - 1)\), must be zero.

\(\alpha - 1 = 0\)

\(\alpha = 1\)

So, the common root of the two quadratic equations is 1.

Finding the Value of a + b

Now that we know the common root \(\alpha = 1\), we can substitute this value back into either of the original equations to find a relationship between \(a\) and \(b\). Let's use Equation 1:

\(x^2 + ax + b = 0\)

Substitute \(x = 1\):

\(1^2 + a(1) + b = 0\)

\(1 + a + b = 0\)

Rearranging the terms to find the value of \(a + b\):

\(a + b = -1\)

Let's check with Equation 2 as well:

\(x^2 + bx + a = 0\)

Substitute \(x = 1\):

\(1^2 + b(1) + a = 0\)

\(1 + b + a = 0\)

\(a + b = -1\)

Both equations give the same result, confirming that the value of \(a + b\) is -1 when the common root is 1.

Final Answer Determination

Based on our calculations, if the two quadratic equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\) have a common root and \(a \neq b\), then the value of \(a + b\) is -1.

Revision Table: Key Steps

Additional Information on Quadratic Equations and Roots

A quadratic equation is a polynomial equation of the second degree. The general form is \(Ax^2 + Bx + C = 0\), where \(A, B, C\) are coefficients and \(A \neq 0\). The solutions for \(x\) are called the roots of the equation.

Methods to Find Roots:

• Factoring: If the quadratic expression can be factored into two linear factors, set each factor to zero to find the roots.
• Completing the Square: A method to rewrite the quadratic expression in the form \((x-h)^2 = k\), which can then be solved by taking the square root.
• Quadratic Formula: The roots of \(Ax^2 + Bx + C = 0\) are given by the formula:

  \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)

The Discriminant:

The term inside the square root in the quadratic formula, \(\Delta = B^2 - 4AC\), is called the discriminant. It tells us about the nature of the roots:

• If \(\Delta > 0\), there are two distinct real roots.
• If \(\Delta = 0\), there is exactly one real root (a repeated root).
• If \(\Delta < 0\), there are two complex conjugate roots.

Vieta's Formulas:

For a quadratic equation \(Ax^2 + Bx + C = 0\) with roots \(r_1\) and \(r_2\), Vieta's formulas relate the coefficients to the roots:

• Sum of roots: \(r_1 + r_2 = -\frac{B}{A}\)
• Product of roots: \(r_1 \times r_2 = \frac{C}{A}\)

In our problem, for \(x^2 + ax + b = 0\), let the roots be 1 and \(r_1\). The sum of roots is \(1 + r_1 = -a\) and the product is \(1 \times r_1 = b\), so \(r_1 = b\) and \(1 + b = -a \implies a + b = -1\). Similarly, for \(x^2 + bx + a = 0\), let the roots be 1 and \(r_2\). The sum of roots is \(1 + r_2 = -b\) and the product is \(1 \times r_2 = a\), so \(r_2 = a\) and \(1 + a = -b \implies a + b = -1\). This confirms our result.