Square roots within an equation?

Question

Help !!!

(7 + sqrt3)(7-sqrt3)


Also I need to learn the algebraic steps in order to solve this problem

(sqrt y + 18 ) = 4

this is a similair problem I am aslo struggling with that I need to know the algebraic steps to

(sqrt5x + 5) -15 = 0

Thanks for your help !

Answer 1

FOIL:

(7 + √3)(7 - √3) = 49 - 7√3 + 7√3 + √3√3 = 49 - 3 = 46

----

(√y + 18) = 4

√y = -14

Square both sides:

y = (-14)² = 196

but when you check...it doesn't work:

√196 + 18 = 14 + 18 = 32...so there's no solution.

UNLESS you meant....

√(y + 18) = 4

Then square both sides first:

y + 18 = 16
y = -2

THAT would work in the original equation.

--------

√5x + 5 - 15 = 0
√5x - 10 = 0
√5x = 10
Square both sides:
5x = 100
x = 20

Unless you meant....

√(5x + 5) - 15 = 0

In which case it would be:

√(5x + 5) = 15
Square both sides:
5x + 5 = 15² = 225
5x = 220
x = 44

Answer 2

Using (a+b)*(a-b) = a^2-b^2 (7+sqrt3)* (7-sqrt3)=49-3=46As you write it sqrty = 4-18 =-14 Normally if ther is no sign in front of a sqrt it´s supposed that the sqrt is positive.
So sqrt y = -14 has no meaning
sqrt(5x)= 15 -5 =10==> 5x = 100 ( squaring both sides) x=20

Answer 3

1.In this case you apply the formula (a+b)(a-b)=a^2-b^2
(7+sqrt3)(7-sqrt3)
=(7)^2-(sqrt3)^2
=49-3
=46
2sqrt y +18=4
=>sqrt y=4-18=-14
=>.y=196 [squaring both sides]
3.sqrt 5x+5-15=0
sqrt5x=15-5=10
=>5x=100 [squaring both sides]
=>x=100/5=20

Answer 4

1. (7 + sqrt.3)(7-sqrt3)
This is a formula :- a^2 - b^2 = (a+b)(a-b)

Therefore :-
the equation becomes :- 7^2 - sqrt3 ^2 = 49-3
= 46