(marking brainliest) please help
asap! questions are in the pdf.

1) x = 6 + 2√19 or x = 6 - 2√19

2a) (x + 5)²

2b) (x + 5/2)²

3) x = 22 or 18

Step-by-step explanation:

1) x² - 12x = 40

First step is;

We take ½ of the x-term and square it.

Thus;

(½ × -12)² = (-6)² = 36

Second step is to add the result in step 1 to both sides;

x² - 12x + 36 = 40 + 36

x² - 12x + 36 = 76

On the left, is a perfect square that can be written as (x - 6)²

Thus, the new equation is;

(x - 6)² = 76

Taking square root of both sides gives;

x - 6 = ±√76

x - 6 = ±2√19

x = 6 + 2√19 or x = 6 - 2√19

2a) we want to find the value of c that makes x² + 10x + c the perfect square.

From completing the square in question 1 above, we see that;

c = (b/2)²

From ax² + bx + c, we can see that in this case, b = 10

Thus;

c = (10/2)²

c = 5²

c = 25

Thus, the trinomial is;

x² + 10x + 25

As a perfect square, it is;

(x + 5)²

2b) x² + 5x + c

Similar to 2a above;

b = 5

Thus; c = (5/2)²

c = 25/4

Trinomial is; x² + 5x + 25/4

As a perfect square, we have;

(x + 5/2)²

3) We want to solve x² - 40x + 396 = 0

This can be rewritten as;

x² - 40x = -396

First step is;

We take ½ of the x-term and square it.

Thus;

(½ × -40)² = 400

Second step is to add the result in step 1 to both sides;

x² - 40x + 400 = -396 + 400

x² - 40x + 400 = 4

On the left, is a perfect square that can be written as (x - 20)²

Thus;

(x - 20)² = 4

Taking square root of both sides gives;

x - 20 = ±√4

x - 20 = ±2

x = 20 + 2 or x = 20 - 2

x = 22 or 18

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)