Key Concepts and Formulas

• For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is given by $D = B^2 - 4AC$.
• The quadratic equation has real roots if and only if $D \ge 0$.
• To solve a quadratic inequality, find the roots of the corresponding quadratic equation and test intervals.

Step-by-Step Solution

Step 1: Rewrite the Equation in Standard Quadratic Form

We need to rewrite the given equation $2x^2 + (a - 10)x + \frac{33}{2} = 2a$ in the standard form $Ax^2 + Bx + C = 0$. Subtracting $2a$ from both sides, we get: $2x^2 + (a - 10)x + \frac{33}{2} - 2a = 0$ Now, we can identify the coefficients:

• $A = 2$
• $B = a - 10$
• $C = \frac{33}{2} - 2a$

Explanation: Putting the equation in standard form is a prerequisite for correctly applying the discriminant formula. It ensures that the coefficients A, B, and C are correctly identified.

Step 2: Apply the Discriminant Condition for Real Roots

For the equation to have real roots, the discriminant must be greater than or equal to zero: $D \ge 0$. Substituting the values of A, B, and C into the discriminant formula $D = B^2 - 4AC$, we have: $(a - 10)^2 - 4(2)\left(\frac{33}{2} - 2a\right) \ge 0$

Explanation: This step translates the condition of real roots into an inequality involving 'a', which we can then solve.

Step 3: Simplify the Inequality

Expanding and simplifying the inequality: $a^2 - 20a + 100 - 8\left(\frac{33}{2} - 2a\right) \ge 0$ $a^2 - 20a + 100 - 4(33) + 16a \ge 0$ $a^2 - 20a + 100 - 132 + 16a \ge 0$ $a^2 - 4a - 32 \ge 0$

Explanation: This step involves algebraic simplification to obtain a standard quadratic inequality. Accuracy in expanding and combining like terms is crucial.

Step 4: Solve the Quadratic Inequality

We need to find the values of $a$ that satisfy $a^2 - 4a - 32 \ge 0$. First, find the roots of the corresponding quadratic equation $a^2 - 4a - 32 = 0$. Factoring the quadratic, we look for two numbers that multiply to -32 and add up to -4. These numbers are -8 and 4. $(a - 8)(a + 4) \ge 0$ The critical points are $a = 8$ and $a = -4$.

Since the parabola $y = a^2 - 4a - 32$ opens upwards, the inequality is satisfied when $a$ is outside the roots. Thus, $a \le -4 \quad \text{or} \quad a \ge 8$

Explanation: Solving the quadratic inequality involves finding the roots and then determining the intervals where the quadratic expression is non-negative.

Step 5: Identify the Least Positive Value of 'a'

We are looking for the least positive value of $a$. From the solution to the inequality, $a \le -4$ or $a \ge 8$. The condition $a \le -4$ includes only negative values. The condition $a \ge 8$ includes all values greater than or equal to 8.

Therefore, the least positive value of $a$ is $8$.

However, the provided "Correct Answer" is 0. Let's re-examine the question and the steps. The question asks for the least positive value of 'a', meaning 'a' must be strictly greater than 0. Our solution a >= 8 fulfills this. If the correct answer is indeed 0, it implies a mistake somewhere. Let's double check each step.

Step 1: Correct. Standard form. Step 2: Correct. Applying the discriminant. Step 3: Correct. Expanding and simplifying. Step 4: Correct. Solving the inequality. Step 5: If a=0, then the equation becomes 2x^2 - 10x + 33/2 = 0. The discriminant is D = (-10)^2 - 4(2)(33/2) = 100 - 132 = -32 < 0. So, a=0 does NOT give real roots.

There seems to be an error in the provided correct answer. The least positive value for 'a' for which the given quadratic equation has real roots is indeed 8.

Common Mistakes & Tips

• Always rewrite the quadratic equation in standard form before identifying A, B, and C.
• Be careful with signs when expanding and simplifying the discriminant inequality.
• Remember to consider the direction of the parabola when solving quadratic inequalities.

Summary

We found the discriminant of the quadratic equation and set it to be greater than or equal to zero to ensure real roots. We then solved the resulting quadratic inequality and identified the least positive value of 'a' that satisfies the inequality. The least positive value of 'a' is 8.

The final answer is \boxed{8}.