If a+b = $\frac{17}{6}$ and ab=2, then what is the value of (a −b), where a > b ?

Correct Answer: $\frac{1}{6}$

Step-by-Step Explanation:

We are given the following information:

$a+b=\frac{17}{6}$

$ab=2$

We need to find the value of $a-b$, given the condition that $a>b$.

Step 1: Use the fundamental algebraic identity
We recall the algebraic identity that connects the sum, product, and difference of two numbers:
$(a-b{)}^2=(a+b{)}^2-4ab$

Step 2: Substitute the known values into the identity
Substituting $a+b=\frac{17}{6}$ and $ab=2$ into the equation:
$(a-b{)}^2={\left(\frac{17}{6}\right)}^2-4(2)$

Step 3: Perform the calculations
First, calculate the square of $\frac{17}{6}$:
${\left(\frac{17}{6}\right)}^2=\frac{{17}^2}{6^2}=\frac{289}{36}$

Next, evaluate the second term:
$4\times 2=8$

Convert 8 to a fraction with a common denominator of 36:
$8=\frac{8\times 36}{36}=\frac{288}{36}$

Now, subtract the two fractions:
$(a-b{)}^2=\frac{289}{36}-\frac{288}{36}=\frac{1}{36}$

Step 4: Solve for $a-b$
Taking the square root of both sides:
$a-b=\sqrt{\frac{1}{36}}=\pm \frac{1}{6}$

Since we are given that $a>b$, the value of $a-b$ must be strictly positive.
Hence:
$a-b=\frac{1}{6}$