Chapter 1.4, Problem 25PPE

To determine

To Evaluate: The expression $\frac{2b+3c^2}{4a^2-2b}$

Answer to Problem 25PPE

  $\frac{2b+3c^2}{4a^2-2b}=3\frac{4}{31}$

Explanation of Solution

Given information:

The expression $\frac{2b+3c^2}{4a^2-2b}$ and value of $a=8,b=4\&c=16$

Concept and Formula Used:

Rule of exponent and addition and subtraction.

  $a^n=a\times a\times a\times \mathrm{..................}n\text{ times}\text{.}$

BODMAS rule

Calculation:

The given expression is $\frac{2b+3c^2}{4a^2-2b}$

Putting the values of $a=8,b=4\&c=16$ in the expression

  $\begin{array}{l}\frac{2b+3c^2}{4a^2-2b}=\frac{2\cdot 4+3\cdot {16}^2}{4\cdot 8^2-2\cdot 4}\\ \frac{2b+3c^2}{4a^2-2b}=\frac{2\cdot 4+3\cdot 256}{4\cdot 64-2\cdot 4}\\ \text{Multiply the powers 2\&4,3\&256,4\&64,2\&4}\\ \frac{2b+3c^2}{4a^2-2b}=\frac{8+768}{256-8}\\ \text{Add 8\&768 in numerator}\\ \frac{2b+3c^2}{4a^2-2b}=\frac{776}{256-8}\\ \text{Subtract 8 from 256 in denominator}\\ \frac{2b+3c^2}{4a^2-2b}=\frac{776}{248}\\ \text{Simplify }\\ \frac{2b+3c^2}{4a^2-2b}=\frac{97}{31}=3\frac{4}{31}\end{array}$

Conclusion:

  $\frac{2b+3c^2}{4a^2-2b}=3\frac{4}{31}$