SOLUTION: Someone please help me by Solving Equations With Radicals, thanks! 1. <a href="http://photobucket.com" target="_blank"><img src="http://i10.photobucket.com/albums/a139/mzzunders

Question 29360: Someone please help me by Solving Equations With Radicals, thanks!
1.

2.

3.
4.
Found 3 solutions by askmemath, sdmmadam@yahoo.com, josmiceli:
Answer by askmemath(368)   (Show Source): You can put this solution on YOUR website!
1.Squaring on both sides
16 = X-2
Adding 2 on both sides
18 = X
2.Dividing by 2 on both sides

Squaring on both sides
X+3 = 25
Subtracting 3 on both sides
X = 22

3.Squaring on both sides

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, -2. Here's your graph:

4.Subtracting 1 on both sides

Squaring on both sides
X = 16
Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!
1.sqrt(x-2) = 4
Squaring both the sides
(x-2) =4^2 (using [sqrt(p)]^2 = p and here p = (x-2) )
x-2 = 16
x = 16+2
x=18
Answer: x= 18
Verification:sqrt(x-2) =sqrt(18-2)= sqrt(16) = 4 which is what is given.
2. 2[sqrt(x+3)] = 10
Squaring both the sides,
4(x+3) =10^2 (using [sqrt(p)]^2 = p and here p = (x+3) )
4x+12 = 100
4x= 100-12
4x=88
x=88/4=22
Answer: x=22
Verification:
2[sqrt(x+3)] =22[sqrt(22+3)]=2[sqrt(25)] =2X(5) = 10 which is correct

3. x = sqrt(x+6)
Squaring both the sides,
x^2 = (x+6) (using [sqrt(p)]^2 = p and here p = (x+6) )
x^2-x-6=0 ----(*)Which is a quadratic in x
x^2-3x+2x-6 =0
(x^2-3x)+(2x-6) =0 (by additive associativity)
x(x-3)+2(x-3)
(x-3)(x+2)= 0
(x-3) =0 implies x=3 and
(x+2)= 0 implies x= -2
Answer: x=3 and x=-2
Verification: We may verify orally and see that both the values hold.
Note:(On the LHS of (*) splitting the middle term two terms
whose sum is the mid term and whose product is the product of the
square term and the constant term.
So here (-x) = (-3x+2x) and (-3x)X(2x) = -6x^2 = (x^2)X(-6)
4. sqrt(x) + 1 = 5
sqrt(x) = 5 - 1
sqrt(x) =4
Squaring both the sides
x= 16 (using [sqrt(p)]^2 = p and here p = (x) )
Answer: x=16
Verification: Very clear orally even!

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!

square both sides

check

this is true- one of the square roots of 16 is 4, the othewr is -4
--------------------

square both sides

check

true, one of the square roots of 25 is +5, the other is -5
------------------

square both sides

solve for x

x = 3
x = -2
check

true- one of the square roots of 9 is +3, the other is -3

true one of the square roots of 4 is -2, the other is +2
----------------

square both sides

check

true- one of the square roots of 16 is 4, the other is -4

RELATED QUESTIONS
Perform the indicated operation and expression lowest terms. Thanks for any help. (answered by CubeyThePenguin)
Can you please help me with Simplifying Rational Expressions?, thank you! 1) (answered by elima)
Can someone pls help me with Simplifying Rational Expressions?, thanks! 1) (answered by sdmmadam@yahoo.com,Earlsdon)
Perform the indicated operation and expression lowest terms. (answered by CubeyThePenguin)
Help me pls.. Multiplying Radicals (answered by Nate)
Please help me with this proof: (answered by jim_thompson5910)
Can someone help me with these 2 questions. (answered by ewatrrr)
Please help me understand this problem aslo please... (answered by funmath)
Please help me solve the equation. (answered by Nate)