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
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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:
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4.Subtracting 1 on both sides
Squaring on both sides
X = 16
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!
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