Question

Find the value of
\[ \frac{(0.6)^4-(0.5)^4}{(0.6)^2+(0.5)^2} \]

• A. \(0.0121\)
• B. \(0.011\)
• C. \(0.11\)
• D. \(1.1\)

Hint:

Use \(a^4-b^4=(a^2-b^2)(a^2+b^2)\) to cancel the denominator.

Answer 1

The Correct Option is C

Step 1: Spot the algebraic identity hiding in the numerator.
The numerator is a difference of fourth powers, \(a^4-b^4\), with \(a=0.6\) and \(b=0.5\). This always factors as
\[ a^4 - b^4 = (a^2-b^2)(a^2+b^2) \]
because \(a^4-b^4\) is itself a difference of squares of \(a^2\) and \(b^2\).

Step 2: Cancel the common factor with the denominator.
The denominator of the given expression is exactly \(a^2+b^2\), which is one of the two factors above. So the whole expression simplifies to
\[ \frac{(a^2-b^2)(a^2+b^2)}{a^2+b^2} = a^2-b^2 \]
Since \(a^2+b^2 = 0.36+0.25=0.61\) is not zero, this cancellation is valid.

Step 3: Factor \(a^2-b^2\) once more and plug in numbers.
\[ a^2-b^2 = (a-b)(a+b) \]
With \(a=0.6\) and \(b=0.5\):
\[ a-b = 0.1, \qquad a+b = 1.1 \]
\[ a^2-b^2 = 0.1 \times 1.1 = 0.11 \]

Step 4: Why the other options are wrong.
Option (a), 0.0121, is \((0.11)^2\), the result of squaring the correct answer by mistake. Option (b), 0.011, is off by a factor of 10, a common decimal-point slip. Option (d), 1.1, is just \(a+b\) on its own, without multiplying by \(a-b\).

Final Answer:
The value of the expression is 0.11. \[ \boxed{0.11} \]