# What are the tw numbers that root (5) lies between them?two numbers in one decimal place

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The two perfect square integers above and below 5 are 9 and 4.

Therefore 4 <5 < 9.

Taking the positive square root, we get:

sqrt4 <sqrt5 < sqrt9

2 < sqrt5 < 3.

Therefore (2+x)^2 = 5.

4+ 4x+x^2 = 5

4x+x^2 = 5-4 = 1.

4x+x^2 = 1

Take x= 0.3 or x = 0.2

4x+x^2 = 4*(.2)+(0.2)^2 = 0.84 < 1 < 4(0.3)+(0.3)^3 =1.29

So now 2.2< sqrt5 < 2.3.

Therefore 2.2^2 < 5 < 2.3^2.

So (2.2+x)^2 = 5.

4.84 +2*2.2x+x^2 < 5.

4.4x +x^2 < 5-4.84 = 0.16.

4.4x+x^2 = 0.16

Take x = 0.03 or x =0.04.

4.4x+x^2 = 4.4(0.03)+0.03)^2 = 0.1329 < 0.16 < 4.4(0.04)+0.04^2 = 0.1776.

We can continue like this to obtain more and more narrower limits within which square root of 5 lies.

Therefore 2.23^2 < 5 < 4.24^2

We take the positive square root :

Therefore 2.23 < sqrt5 < 2.24

Now we can approximate the square root of 5 to the first decimal by writing the inequality:

2.2 < sqrt5 < 2.3.