Question

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

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

Answer 1

The correct option is B

$\frac{1}{4}$

$49a^2-b=(7a{)}^2-(\sqrt{b}{)}^2=(7a+\sqrt{b})(7a-\sqrt{b})\therefore (7a+\sqrt{b})(7a-\sqrt{b})=(7a+\frac{1}{2})(7a-\frac{1}{2})\text{Comparing,}\sqrt{b}=\frac{1}{2}\Rightarrow b=\frac{1}{4}$