Which of the following quadratic equations has 3 as a root?

Introduction / Context:
This question checks whether you can verify if a given number is a root of a quadratic equation. It is a basic skill in algebra, often used when factoring quadratics or solving multiple choice questions quickly.

Given Data / Assumptions:

• Candidate root: x = 3.
• Four quadratic equations are provided.
• We need to find which equation is satisfied by x = 3.

Concept / Approach:
To test whether a number is a root of an equation, we substitute that number into the left hand side of the equation and check whether the result is zero. If the value is zero, the number is a root. If not, it is not a root. We will apply this check to each option.

Step-by-Step Solution:
Step 1: Check option (a): x^2 - 7x + 12 = 0 at x = 3.Step 2: Compute 3^2 - 7*3 + 12 = 9 - 21 + 12.Step 3: 9 - 21 = -12 and -12 + 12 = 0, so x = 3 satisfies equation (a).Step 4: Check option (b): x^2 - 5x + 12 at x = 3 gives 9 - 15 + 12 = 6, not zero.Step 5: Check option (c): x^2 + 3x + 12 at x = 3 gives 9 + 9 + 12 = 30, not zero.Step 6: Check option (d): x^2 + 3x - 12 at x = 3 gives 9 + 9 - 12 = 6, not zero.Step 7: Only option (a) yields zero, so it is the correct equation.

Verification / Alternative check:
We can factor equation (a): x^2 - 7x + 12 = 0. This factors as (x - 3)(x - 4) = 0, so the roots are x = 3 and x = 4. This confirms that x = 3 is indeed a root of this equation.

Why Other Options Are Wrong:

• x^2 - 5x + 12 = 0: Substituting x = 3 gives 6, not 0.
• x^2 + 3x + 12 = 0: Substituting x = 3 gives 30, not 0.
• x^2 + 3x - 12 = 0: Substituting x = 3 gives 6, not 0.

Common Pitfalls:
Some candidates may try to factor each quadratic blindly, which takes more time than simply substituting the given value. Others may make simple arithmetic mistakes while evaluating the expressions at x = 3, so careful calculation is essential.

Final Answer:
The equation that has 3 as a root is x^2 - 7x + 12 = 0.