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

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Question 29360: Someone please help me by Solving Equations With Radicals, thanks!
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Found 3 solutions by askmemath, sdmmadam@yahoo.com, josmiceli:
Answer by askmemath(368) About Me  (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
sqrt%28X%2B3%29+=+5
Squaring on both sides
X+3 = 25
Subtracting 3 on both sides
X = 22

3.Squaring on both sides
X%5E2+=+X%2B6
X%5E2+-+X-6+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-6+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%28-1%29-sqrt%28+25+%29%29%2F2%5C1+=+-2

Quadratic expression 1x%5E2%2B-1x%2B-6 can be factored:
1x%5E2%2B-1x%2B-6+=+1%28x-3%29%2A%28x--2%29
Again, the answer is: 3, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-6+%29


4.Subtracting 1 on both sides
sqrt%28X%29+=+4
Squaring on both sides
X = 16

Answer by sdmmadam@yahoo.com(530) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x+-+2%29+=+4
square both sides
x+-+2+=+16
x+=+18
check
sqrt%28x+-+2%29+=+4
sqrt%2818+-+2%29+=+4
sqrt%2816%29+=+4
this is true- one of the square roots of 16 is 4, the othewr is -4
--------------------
2+%2A+sqrt%28x+%2B3%29+=+10
square both sides
4+%2A+%28x+%2B+3%29+=+100
x+%2B+3++=+100+%2F+4
x+=+25+-+3
x+=+22
check
2+%2A+sqrt%28x+%2B3%29+=+10
2+%2A+sqrt%2822+%2B3%29+=+10
sqrt%2825%29+=+10%2F2
5+=+5
true, one of the square roots of 25 is +5, the other is -5
------------------
x+=+sqrt%28x+%2B+6%29
square both sides
x%5E2+=+x+%2B+6
x%5E2+-+x+-+6+=+0
solve for x
%28x+-+3%29+%2A+%28x+%2B+2%29+=+0
x = 3
x = -2
check
x+=+sqrt%28x+%2B+6%29
3+=+sqrt%283+%2B+6%29
3+=+sqrt%289%29
true- one of the square roots of 9 is +3, the other is -3
-2+=+sqrt%28-2+%2B+6%29
-2+=+sqrt%284%29
true one of the square roots of 4 is -2, the other is +2
----------------
sqrt%28x%29+%2B+1+=+5
sqrt%28x%29+=+4
square both sides
x+=+16
check
sqrt%28x%29+%2B+1+=+5
sqrt%2816%29+%2B+1+=+5
4+%2B+1+=+5
true- one of the square roots of 16 is 4, the other is -4