So, the roots of the equation are the values of x for which
`( sqrt3x-sqrt2) ( sqrt3x-sqrt2) = 0`
Now `sqrt3x-sqrt2 = 0` for `x = sqrt(2/3)`
So, this root is repeated twice, one for each repeated factor `sqrt3x-sqrt2`.
Therefore, the roots of `3x^2-2 sqrt6x+2 = 0` are `sqrt(2/3) , sqrt(2/3)`
Find the dimensions of the prayer hall discussed in a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall ? Class 10 Chapter 4 Example 6
Suppose the breadth of the hall is `x` metres. Then,
its length should be `(2x + 1)` metres. We can depict
this information pictorially as shown in Fig.
Now, area of the hall` = (2x + 1). x m^2 = (2x^2 + x) m^2`
So, `2x^2 + x = 300` (Given)
Therefore, `2x^2 + x – 300 = 0`
Applying the factorisation method, we write this equation as
`2x^2 – 24x + 25x – 300 = 0`
`2x (x – 12) + 25 (x – 12) = 0`
i.e.,` (x – 12)(2x + 25) = 0`
So, the roots of the given equation are `x = 12` or `x = – 12.5.` Since `x` is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is `12 m`. Its length `= 2x + 1 = 25 m.`