How to rewrite y = x^2 – 20x + 111 into four terms in which the first three terms form a perfect square trinomial

x^2 - 20x + 100 + 11 = (x - 10)^2 + 11

Step-by-step explanation:

We're basically completing the square here.

We have the equation x^2 - 20x + 11. Let's just focus on the first two terms for now:

x^2 - 20x

In order to make this a perfect square trinomial, we need to find the constant term. We use the formula (b/2)^2, where b is the coefficient of the x term. Here, b = -20, so the constant term will be: (-20/2)^2 = (-10)^2 = 100.

We now have: x^2 - 20x + 100

However, we can't forget about the 111 from the original equation. We can't do this: x^2 - 20x + 100 + 111 because then we're left with an equation that is NOT the same as the original. Instead, we need to subtract 100 from 11 because we're already adding it in our perfect square trinomial:

x^2 - 20x + 100 + 111 - 100

x^2 - 20x + 100 + 11

The equation can then be written as: (x - 10)^2 + 11

Hope this helps!

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl