# Linear Approximation (Need someone to check my work)

#### shamieh

##### Active member
Use a linear approximation to find a good approximation to $$\displaystyle \sqrt{100.4}$$

$$\displaystyle x = 100.4$$
$$\displaystyle x1 = 100$$
$$\displaystyle y1 = 10$$

$$\displaystyle y - 10 = \frac{1}{20}(100.4 - 100)$$

$$\displaystyle y = 10.20$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Use a linear approximation to find a good approximation to $$\displaystyle \sqrt{100.4}$$

$$\displaystyle x = 100.4$$
$$\displaystyle x1 = 100$$
$$\displaystyle y1 = 10$$

$$\displaystyle y - 10 = \frac{1}{20}(100.4 - 100)$$

$$\displaystyle y = 10.20$$
Looks correct... except for a small calculation mistake.
Did you check what $10.20^2$ is?
That should immediately reveal the mistake.

#### shamieh

##### Active member
Looks correct... except for a small calculation mistake.
Did you check what $10.20^2$ is?
That should immediately reveal the mistake.
its 104.04 but I don't understand where I went wrong. Why can't I say 1/20 = .05 and then say .05 * .4 = .20, then finally 10 + .20 = 10.20 ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
its 104.04 but I don't understand where I went wrong. Why can't I say 1/20 = .05 and then say .05 * .4 = .20, then finally 10 + .20 = 10.20 ?
As you can see your fraction is off by a factor of 10.
Indeed .05 * .4 ≠ .20.
Instead .05 * .4 = .020.

The trick is to count the number of digits after the decimal point.
.05 has 2 digits, .4 has 1 digit, therefore their product (5 x 4 = 20) must have 2+1=3 digits after the decimal point (0.020).

#### shamieh

##### Active member
10.02 is the correct answer then correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
10.02 is the correct answer then correct?
Let's see:
$$10.02^2 = (10 + 0.02)^2 = 10^2 + 2 \cdot 10 \cdot 0.02 + 0.02^2 = 100 + 0.4 + 0.0004 = 100.4004$$
Yep. I'd say that's the correct answer.