Question

Find the zeroes of the polynomial $2x^2 - 6$.

Hint:

For equations of type $ax^2 - b = 0$:
Move constant term to RHS.
Divide by coefficient of $x^2$.
Take square root.
Always include both $+$ and $-$ roots.

Answer 1

Step 1: Understanding the Question:
The question asks for the "zeroes" of the polynomial \(2x^2 - 6\). The zeroes (or roots) of a polynomial are the values of the variable \(x\) for which the polynomial evaluates to zero.
Step 2: Key Formula or Approach:
To find the zeroes of any polynomial \(P(x)\), we set the polynomial equal to zero, i.e., \(P(x) = 0\), and then solve the resulting equation for \(x\).
Step 3: Detailed Explanation:
The given polynomial is \(P(x) = 2x^2 - 6\).
Set the polynomial equal to zero:
\[ 2x^2 - 6 = 0 \] To solve for \(x\), we first need to isolate the \(x^2\) term. We can do this by adding 6 to both sides of the equation:
\[ 2x^2 = 6 \] Next, divide both sides by the coefficient of \(x^2\), which is 2:
\[ x^2 = \frac{6}{2} \] \[ x^2 = 3 \] Finally, to find \(x\), we take the square root of both sides. It is important to remember that taking a square root yields both a positive and a negative result.
\[ x = \pm \sqrt{3} \] Thus, the two zeroes of the polynomial are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
Step 4: Final Answer:
The zeroes of the polynomial \(2x^2 - 6\) are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
\[ \boxed{x = \pm \sqrt{3}} \]