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

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

That is a fantastic question! The problem asks for 'real roots'. In mathematics, 'real roots' includes both distinct real roots ($D > 0$) and equal real roots ($D = 0$). If the question had asked for 'distinct real roots', we would strictly use $D > 0$. Since it just says 'real roots', we must include the case where the roots are equal, hence $D \ge 0$.

Q: Why did we have to move $2a$ to the left side? Can't we just use the equation as it is?

Think of the standard quadratic form $Ax^2 + Bx + C = 0$ as the 'DNA' of the quadratic equation. The discriminant formula $D = B^2 - 4AC$ is specifically designed to work with this structure. If you don't bring all terms to one side, your constant term $C$ will be incorrect, leading to a completely wrong discriminant. Always 'dress' your equation in standard form before you start!

Q: I got the intervals $(-\infty, -4] \cup [8, \infty)$. Why is the answer 8 and not -4?

You are almost there! The question specifically asks for the 'least positive value' of 'a'. While both $-4$ and $8$ are boundary points of our valid intervals, $-4$ is a negative number. The positive values of 'a' start from $8$ and go up to infinity. Therefore, the smallest positive value in that set is indeed $8$.

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

Welcome, fellow traveler in the realm of mathematics! Today, we are going to tackle a problem that might look like a simple algebraic exercise, but it is actually a beautiful test of your foundational understanding of quadratic equations.

We are given the equation $2x^2 + (a - 10)x + \frac{33}{2} = 2a$ and told that it has real roots. Our mission is to find the least positive value of the parameter $a$.

### The Art of Standardization

Before we can do anything, we must respect the structure of the quadratic equation. The standard form, $Ax^2 + Bx + C = 0$, is our North Star.

Looking at our given equation, $2x^2 + (a - 10)x + \frac{33}{2} = 2a$, we see that the right side is cluttered with a $2a$. By subtracting $2a$ from both sides, we transform our equation into:

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

Now, it is perfectly dressed. We can clearly see that $A = 2$, $B = (a - 10)$, and our constant term $C = (\frac{33}{2} - 2a)$. This clarity is the secret to avoiding those frustrating 'silly mistakes' that haunt so many exams.

### The Golden Rule of Real Roots

What does it mean for a quadratic to have real roots? It means the parabola either cuts the $x$-axis at two points or just touches it at one.

Mathematically, this is governed by the discriminant, $D = B^2 - 4AC$. For real roots, we require $D \ge 0$. This is our golden rule. Let's substitute our identified coefficients into this inequality:

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

### The Algebraic Grind

Let's simplify this piece by piece. First, the expansion of $(a - 10)^2$ is a classic identity: $a^2 - 20a + 100$.

Now, for the second part: $-4(2)(\frac{33}{2} - 2a)$. This becomes $-8(\frac{33}{2} - 2a)$, which simplifies to $-132 + 16a$. Putting it all together, our inequality becomes:

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

Combining the like terms, $-20a + 16a$ gives us $-4a$, and $100 - 132$ gives us $-32$. We are left with a beautifully simple quadratic inequality:

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

### The Wavy Curve

We are almost at the finish line. We need to factorize $a^2 - 4a - 32$. We are looking for two numbers that multiply to $-32$ and add up to $-4$.

A quick mental check reveals $-8$ and $+4$. So, our inequality is:

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

Using the wavy curve method, we plot the critical points $a = -4$ and $a = 8$ on a number line. Since the quadratic opens upwards, the expression is positive outside the roots. Thus, the valid intervals are $a \in (-\infty, -4] \cup [8, \infty)$.

### Final Calculation

The question asks for the **least positive value** of $a$. Looking at our intervals, the positive values start from $8$ and extend to infinity.

The smallest value in this set is clearly **$8$**. Through careful standardization, discriminant application, and algebraic simplification, we have arrived at the truth.

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

Condition for Real Roots

Standard Quadratic Form

Applying $D \geq 0$

Simplifying the Constant Part

Expanding $(a - 10)^{2}$

Forming the Final Inequality

Factorizing the Quadratic

Plotting Critical Points

Identifying the Valid Intervals