Question

If \(49x^2 - b = (7x + \frac{1}{2})(7x - \frac{1}{2})\), then the value of b is :

• A. \(0\)
• B. \(\frac{1}{\sqrt{2}}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{2}\)

Hint:

Recognize the pattern \((A+B)(A-B)\) on the right side. Here, \(A = 7x\) and \(B = 1/2\). The identity is \((A+B)(A-B) = A^2 - B^2\). So, \((7x + \frac{1}{2})(7x - \frac{1}{2}) = (7x)^2 - (\frac{1}{2})^2 = 49x^2 - \frac{1}{4}\). The original equation is \(49x^2 - b = 49x^2 - \frac{1}{4}\). Comparing the two sides, it's clear that \(b = \frac{1}{4}\).

Answer 1

The Correct Option is C

Concept: This problem uses the algebraic identity for the difference of two squares: \((A+B)(A-B) = A^2 - B^2\). Step 1: Identify the pattern on the right-hand side (RHS) of the equation The right-hand side of the equation is \((7x + \frac{1}{2})(7x - \frac{1}{2})\). This is in the form \((A+B)(A-B)\), where:
\(A = 7x\)
\(B = \frac{1}{2}\) Step 2: Apply the difference of squares identity Using the identity \((A+B)(A-B) = A^2 - B^2\): \[ (7x + \frac{1}{2})(7x - \frac{1}{2}) = (7x)^2 - \left(\frac{1}{2}\right)^2 \] Step 3: Simplify the squared terms
\((7x)^2 = 7^2 \times x^2 = 49x^2\)
\(\left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}\) So, the RHS becomes: \[ 49x^2 - \frac{1}{4} \] Step 4: Equate the given equation with the simplified RHS The given equation is \(49x^2 - b = (7x + \frac{1}{2})(7x - \frac{1}{2})\). We found that the RHS simplifies to \(49x^2 - \frac{1}{4}\). Therefore, we have: \[ 49x^2 - b = 49x^2 - \frac{1}{4} \] Step 5: Solve for b For this equality to hold true for all values of \(x\), the corresponding terms must be equal. Comparing the constant terms on both sides (or subtracting \(49x^2\) from both sides): \[ -b = -\frac{1}{4} \] Multiplying both sides by -1: \[ b = \frac{1}{4} \] The value of b is \(\frac{1}{4}\). This matches option (3).