Simran has tried to approximate the solution to x = √40 to 1 d.p.
Her workings are shown below.
a) Identify the mistake Simran has made and explain what she should do instead.
b) Find the approximate solution to x = √40 to 1 d.p.

6² = 36
7² = 49
so √40 is between 6 and 7
so x is 6.5 to 1 d.p.​