Finding the roots by completing the square

1 answer:

6 0

For y=x^2-6x-11

complete the square

so roots
set y=0
0=x^2-6x-11

group x terms
0=(x^2-6x)-11
take 1/2 of linear coefient (-6) and square it
-6/2=-3, (-3)^2=9
add positive and neative to inside of parenthasees
0=(x^2-6x+9-9)-11
complete square
0=((x-3)^2-9)-11
expand
0=(x-3)^2-9-11
0=(x-3)^2-20
add 20 to both sides
20=(x-3)^2
sqrt both sides
remember positive and negative roots

+/-2√5=x-3
add 3 to both sides
3+/-2√5=x

so
x=3+2√5 and 3-2√5

You might be interested in

Plz help lol thx.....

Nataliya [291]

Answer:

Step-by-step explanation:

the domain is the set of numbers that can be x values. in this case, the equation is linear so any number you plug in for x will give you a y value

5 0

3 years ago

The difference between the observed value of the dependent variable and the value predicted using the estimated regression equat

Elenna [48]

Answer:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

$(x_1, y_1) ,....,(x_n,y_n)$

For this case we assume that we want to find a linear regression model given by:

$\hat y = \hat m x +\hat b$

Where:

$\hat m$ represent the estimated slope for the model

$\hat b$ represent the estimated intercept for the model

And for any estimation of the dependent variable $\hat y_i , i=1,...,n$ is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

$e_i = y_i -\hat y_i , i=1,...,n$

So the best option for this case is:

d. residual

Step-by-step explanation: