# Prove that 4*[(sqrt68)-8] < 1.

### 3 Answers | Add Yours

We'll consider the function f(x) = sqrt x and the interval [64,68].

To prove the given inequality, we'll use Lagrange's theoreme.

According to Lagrange's theoreme, for a function f(x), over the interval [a,b]:

f(b)-f(a)=f'(c)(b-a), where c belongs to (a,b)

We'll put a = 64 and b = 68.

sqrt68-sqrt64=(68-64)/(2sqrt c)

sqrt68 - 8=2/sqrt c

If c=64, sqrt 64=8, then 2/sqrt64=2/8=1/4

64<65, sqrt 64<sqrt65; 1/sqrt 64>1/sqrt 65,

2/sqrt64>2/sqrt 65

1/4>2/sqrt 65

So, 1/4 > sqrt68 - 8 => 1>4*(sqrt68 - 8)

To prove 4{(68^(1/2) -8} < 1.

We rationalise numerator by multiplying and dividing by (68^(1/2) -8 )

LHS = 4 (68 ^1/2 -8)*(68^(1/2 +8) / (68^(1/2) +8)

LHS < 4(68-64) / (68^1/2 + 8)

LHS < 4*4/(68^1/2) +8) < 16/ { 64^1/2 +8) , as 68^1/2 is replaced by 64^1/2 which diminished denominator value.

LHS < 16/(8+8) =1

LHS < 1 which proves the problem

4*[(sqrt68)-8] < 1.

L.H.S=4*[(sqrt68)-8]

=4*(8.246211251-8)

=4*0.246211251

=0.984845005

so L.H.S<1