If x=7+root 40 ,find the value of root x+1by the root of x

Chapter : Algebra

Study : Math in Junior high school

x = 7 + √40

find √x of √x + 1

= √x + 1

= √(7+√40) + 1

in Formula is :

= √7+√40 = √x + √y

= (√7+√40)² = (√x + √y)²

= 7+√40 = x + 2√xy + y

= 7 + √40 = x + y + 2√xy

→ 7 = x + y → y = 7 - x ... Equation 1

→ √40 = 2√xy → √40 = 2.2√10 = 4√10

= xy = 10 ... Equation 2

substitution Equation 1 to 2 :

= xy = 10

= x(7-x) = 10

= 7x - x² = 10

= x² - 7x + 10 = 0

= (x - 5)(x - 2) = 0

= x = 5 or x = 2

Subsitution x = 5 and x = 2, to equation 1

#For x = 5

= y = 7 - x

= y = 7 - (5)

= y = 2

#For x = 2

= y = 7 - x

= y = 7 - (2)

= y = 5

and his x and y was find :

#Equation 1 :

= x = 5 and y = 2

#Equation 2 :

= x = 2 and y = 5

So that :

√7+√40 = √x + √y

= √7+√40 = √2 + √5

And that is answer of question :

= √2 + √5 + 1

im sorry but no. i just need to answer for 50 poonts oof

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: