The least positive value of a for which the equation, 2x² + (a - 10)x + 33/2 = 2a has real roots is

Q: Wait, why do we set $D \ge 0$ and not just $D > 0$?

That is a fantastic catch! The problem asks for 'real roots.' In mathematics, real roots include both distinct real roots (where $D > 0$) and equal real roots (where $D = 0$). If we only used $D > 0$, we would miss the case where the parabola just kisses the x-axis at a single point. Always include the equality for 'real roots'!

Q: I got $a^2 - 4a - 32 \le 0$ instead of $\ge 0$. Did I mess up the algebra?

It is very easy to lose a negative sign when distributing $-4AC$. Remember that $C$ is $(\frac{33}{2} - 2a)$. When you multiply $-4 \times 2 \times (-2a)$, you get $+16a$. If you got a different sign, re-check your distribution of the $-8$ factor. Precision is the hallmark of a JEE topper!

Q: Why do we reject the negative interval $(-\infty, -4]$?

The question specifically asks for the 'least positive value' of $a$. While the inequality holds true for values like $-5$ or $-10$, those are negative numbers. By imposing the constraint $a > 0$, we filter out the negative solution set and focus only on the positive domain.

Q: Can I solve this by completing the square instead of using the discriminant?

Absolutely! Completing the square is a powerful tool. You could rewrite the equation as $2(x + \frac{a-10}{4})^2 + (\frac{33}{2} - 2a) - \frac{(a-10)^2}{8} = 0$. For real roots, the constant term must be non-positive. It leads to the exact same inequality. Both methods are elegant; choose the one where you feel most confident.

Q: Is there a shortcut to finding the least value once I have the inequality?

Once you have $(a - 8)(a + 4) \ge 0$, you know the roots are $8$ and $-4$. Since the parabola $f(a) = a^2 - 4a - 32$ opens upward, it is positive outside the roots. The smallest positive value is simply the rightmost root of the quadratic expression, which is $8$.

Question

The least positive value of $a$ for which the equation, $2x^2 + (a - 10)x + \frac{33}{2} = 2a$ has real roots is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex machine—a quadratic equation $2x^2 + (a - 10)x + \frac{33}{2} = 2a$. It has a variable $a$ hiding in the coefficients, shifting the very shape of the parabola as $a$ changes.

Our mission is to find the **least positive value** of $a$ that allows this machine to produce **real roots**. In the world of mathematics, real roots represent points where the system balances or intersects the axis.

### Bringing Order to Chaos

Before we can analyze the roots, we must bring the equation into the standard form $Ax^2 + Bx + C = 0$. We move the $2a$ to the left side to isolate the zero:

$$2x^2 + (a - 10)x + \left( \frac{33}{2} - 2a \right) = 0$$

Now, the coefficients stand revealed: $A = 2$, $B = (a - 10)$, and $C = (\frac{33}{2} - 2a)$. These coefficients dictate the behavior of the parabola.

### The Gatekeeper of Reality

To ensure the roots are real, we invoke the **Discriminant**, $D = B^2 - 4AC$. For real roots, we must satisfy the condition $D \ge 0$.

Substituting our coefficients into the discriminant formula, we obtain:

$$(a - 10)^2 - 4(2)\left( \frac{33}{2} - 2a \right) \ge 0$$

### The Algebraic Crucible

Expanding the term $(a - 10)^2$ yields $a^2 - 20a + 100$. Distributing the constant $-8$ across the term $(\frac{33}{2} - 2a)$ results in $-132 + 16a$.

Combining these components, we arrive at the following inequality:

$$a^2 - 20a + 100 - 132 + 16a \ge 0$$

Simplifying the expression leads to the clean quadratic inequality:

$$a^2 - 4a - 32 \ge 0$$

### The Wavy Curve and the Final Reveal

We factor the quadratic inequality as follows:

$$(a - 8)(a + 4) \ge 0$$

The critical points are $a = 8$ and $a = -4$. Using the wavy curve method, we identify that the inequality holds for $a \in (-\infty, -4] \cup [8, \infty)$.

Since the problem specifically demands the **least positive value** of $a$, we ignore the negative interval. We focus on the interval $[8, \infty)$, where the smallest value is clearly **$8$**.

### Conclusion

When you reach $a = 8$, you have navigated through the algebraic fog to find the exact point where the system transitions into the domain of real roots. This is the essence of mathematical problem solving—transforming a complex expression into a clear, logical boundary. You have mastered the discriminant and emerged with the final result: **$a = 8$**.

Animated Solution for Mathematics - Quadratic Equations: The least positive value of $a$ for which the equation, $2 x^{2} + (a - 10) x + \frac{33}{2} = 2 a$ has real roots is

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Convert to Standard Form

Condition for Real Roots

Substitute Coefficients into $D$

Expand the Squared Term

Simplify the Constant Product

Distribute and Expand Brackets

Combine Like Terms

Factorize the Quadratic

Plot Critical Points

Solve the Inequality

Apply the Positive Constraint

Find the Least Positive Value

The Sigma Insight: Nature of Roots

Analyzing the Setup

Bringing Order to Chaos

The Gatekeeper of Reality

The Algebraic Crucible

The Wavy Curve and the Final Reveal

Conclusion

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least positive value of a for which the equation, 2x2+(a−10)x+233​=2a has real roots is