Question

If there is a common positive root for the equations \(\mathrm{x}^{2}-\mathrm{x}-6=0\) and \(\mathrm{ax}^{2}+3 \mathrm{x}+9=0\), the value of \(\mathrm{a}\) is (a) \(-2\) (b) 2 (c) 3 (d) \(-3\)

Short answer

Question: Given the following equations, find the value of "a" if they have a common positive root: 1) \(x^2 - x - 6 = 0\) 2) \(ax^2 + 3x + 9 = 0\) a) -2 b) -1 c) 1 d) 2 Answer: a) -2

Step-by-step solution

Find the roots of the first equation

To find the roots of the equation \(x^2 - x - 6 = 0\), we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-1\), and \(c=-6\). Plugging these values in the formula, we get: \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}\) \(x = \frac{1 \pm \sqrt{25}}{2}\) The two possible roots are: \(x_1 = \frac{1 + 5}{2} = 3\) \(x_2 = \frac{1 - 5}{2} = -2\)

Find the common positive root

Now, we need to check which of these roots also satisfy the second equation, \(ax^2 + 3x + 9 = 0\). Since we are asked to find a common positive root, we will only check the root \(x_1 = 3\): \(a(3^2) + 3(3) + 9 = 0\) \(9a + 9 + 9 = 0\)

Solve for "a"

Now, we can solve for the value of "a" in the second equation: \(9a + 18 = 0\) \(9a = -18\) \(a = -2\) The value of \(a\) is \(-2\), so the correct answer is (a).

Key concepts

Common Roots

Common roots represent values that satisfy more than one equation simultaneously. In the context of quadratic equations, a common root is one that satisfies both quadratic equations given in a problem. It is crucial when solving problems involving multiple quadratic equations to check if at least one root is shared.

• If two quadratic equations have a common root, you can substitute this root into both equations to find specific parameters or coefficients within the equations, as demonstrated with solving for the value of \( a \).
• Common roots are a hidden intersection point for the two parabolas represented by these equations, meaning they cross the x-axis at the same point.

Identifying common roots can help in simplifying and solving systems involving quadratics, and offer insights into geometric relationships between their graphs.

Quadratic Formula

The quadratic formula is a fundamental tool for solving quadratic equations, expressed as \( ax^2 + bx + c = 0 \). The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]It allows us to find solutions (roots) of any quadratic equation efficiently. Here’s a breakdown of its components:

• \( b^2 - 4ac \) is known as the discriminant. It indicates the nature of the roots: real and distinct, real and equal, or complex.
• The \( \pm \) symbol signifies that there are generally two solutions, corresponding to the two intersections of a parabola with the x-axis.

By substituting values for \( a \), \( b \), and \( c \), you can use the quadratic formula to find the roots of any quadratic equation, as done in the original problem where the formula yielded roots of \( x = 3 \) and \( x = -2 \).

Roots of Equations

Roots of a quadratic equation are values of \( x \) which satisfy the equation such that \( ax^2 + bx + c = 0 \). The concept of roots is fundamental because these points are where the graph of the quadratic intersects the x-axis.

• For an equation \( x^2 - x - 6 = 0 \), solving it results in roots \( x_1 = 3 \) and \( x_2 = -2 \).
• The process of finding roots involves techniques like factoring, completing the square, or using the quadratic formula.
• Roots can be real or complex, and their nature is determined by the discriminant \( b^2 - 4ac \).

Understanding the roots is crucial for any problem involving quadratic equations since they represent the solution set of the equation.

Solving Quadratic Equations

Solving quadratic equations requires identifying the roots, or solutions, that make the equation true. There are several approaches to solve these equations:

• Factoring: This involves expressing the quadratic as a product of two binomials if possible. Not all quadratics can be factored easily.
• Completing the Square: This technique transforms the equation into a perfect square trinomial, making it easier to solve.
• Quadratic Formula: Often the go-to method, as it provides a systematic way to find solutions regardless of whether the equation is easily factorable.

The approach used depends on the specific form of the quadratic equation. In the exercise, the quadratic formula was employed to determine the roots efficiently. With a systematic approach, solving these equations becomes straightforward.