Question

find the values of a and b from the following equality: 68-48√2/4=a+b√2

Answer 1

Given : $68-\dfrac{48\sqrt{2}}{4}=a+b\sqrt2$

To find : The values of a and b.

Step-by-step explanation:

We can re-write the left -side of the given equality as

$68-12\sqrt2$         [∵ 48÷ 4=12]

Now comparing it to the right side of the equality , we get

$68-12\sqrt2=a+b\sqrt2$

It implies

$a=68\quad\quad\text{[Terms without }\sqrt{2}],\\\ b=-12\quad\quad\text{[Coefficient of term having}\sqrt{2}]$

Therefore , the value of a= 68 and the value of b = -12.