Find the two real roots of the quadratic equation x^2 + 3x − 154 = 0, giving the values of x that satisfy this equation.

Introduction / Context:
This problem is a standard quadratic equation question from algebra. You are asked to find the roots of a quadratic polynomial, which means finding the values of x that make the expression x^2 + 3x − 154 equal to zero. Such questions test basic skills in factorisation or use of the quadratic formula.

Given Data / Assumptions:

• Quadratic equation: x^2 + 3x − 154 = 0.
• We are looking for real roots.
• Standard algebraic methods such as factorisation or the quadratic formula apply.

Concept / Approach:
A quadratic equation of the form ax^2 + bx + c = 0 can be solved either by factorisation, by completing the square, or by using the quadratic formula x = [−b ± √(b^2 − 4ac)] / (2a). In this case, the coefficients are simple integers and factorisation is efficient: we search for two numbers whose product is −154 and whose sum is 3.

Step-by-Step Solution:
Step 1: Identify coefficients: a = 1, b = 3, c = −154. Step 2: We need two integers m and n such that m * n = −154 and m + n = 3. Step 3: The factor pairs of 154 are: 1 and 154, 2 and 77, 7 and 22, 11 and 14. Step 4: To get a product of −154, one factor must be negative. Check 11 and −14: 11 * (−14) = −154 and 11 + (−14) = −3, which is not 3. Step 5: Reverse the signs: −11 and 14 give (−11) * 14 = −154 and (−11) + 14 = 3. This pair works. Step 6: Rewrite the quadratic as x^2 + 3x − 154 = (x − 11)(x + 14) = 0. Step 7: Set each factor equal to zero: x − 11 = 0 or x + 14 = 0. Step 8: Solve to obtain x = 11 and x = −14.

Verification / Alternative check:
Substitute x = 11 into the original equation: 11^2 + 3 * 11 − 154 = 121 + 33 − 154 = 154 − 154 = 0. This satisfies the equation. Substitute x = −14: (−14)^2 + 3 * (−14) − 154 = 196 − 42 − 154 = 154 − 154 = 0. Both values satisfy the quadratic, confirming that they are correct roots.

Why Other Options Are Wrong:
11, 14 gives one correct root and one incorrect root because 14 does not satisfy the equation.
14, −11 is just a reordered incorrect pair, since the actual roots are 11 and −14.
14, −22 and 7, −22 do not satisfy the equation when substituted into x^2 + 3x − 154, so they cannot be the correct roots.

Common Pitfalls:
A common error is mixing up the signs of the factor pair. Because the constant term is negative, the factors must have opposite signs. Forgetting to check both possible sign assignments can lead to the wrong pair. Another pitfall is not verifying the roots by substitution, which is a quick and reliable check.

Final Answer:
The roots of the equation are x = 11 and x = −14.