If you have the sqrt(3+x) -1 = sqrt(3x+10), how would you solve for x
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If you have the sqrt(3+x) -1 = sqrt(3x+10), how would you solve for x

[From: ] [author: ] [Date: 12-04-09] [Hit: ]
0sqrt(3 * (-2) + 10) =>sqrt(-6 + 10) =>sqrt(4) =>-2 , 2Once again, if negative roots are permitted, then x = -2 is an answer.If not, then it is extraneous.......

x = -3

sqrt(3 - 3) - 1 =>
0 - 1 =>
-1

sqrt(3 * -3 + 10) =>
sqrt(-9 + 10) =>
sqrt(1) =>
-1 or 1.

x = -3 CAN work, but ONLY if we're allowed to use negative roots. If not, then x = -3 is an extraneous answer


x = -2

sqrt(3 - 2) - 1 =>
+/- 1 - 1 =>
-2 , 0

sqrt(3 * (-2) + 10) =>
sqrt(-6 + 10) =>
sqrt(4) =>
-2 , 2

Once again, if negative roots are permitted, then x = -2 is an answer. If not, then it is extraneous.

But if both answers are extraneous, then what does that mean? That there isn't an answer.


Now, do you remember that I said I could use substitution to solve? Would you like to see how?

-sqrt(3 + x) = x + 3


Let x + 3 = t

-sqrt(t) = t

Let t^(1/2) = u

-u = u^2
0 = u^2 + u
0 = u * (u + 1)

Using the Zero-Product property, we can split this apart (if a * b = 0, then a = 0 or b = 0 , or both a and b are equal to 0)

u = 0

u + 1 = 0
u = -1

u = 0 , -1

u = t^(1/2)

Let's back-substitute

t^(1/2) = 0 , -1
t = 0^2 , (-1)^2
t = 0 , 1
x + 3 = 0 , 1
x = 0 - 3 , 1 - 3
x = -3 , -2

Same possible answers, a slightly different method.

-
Square both sides
(Sqrt(3+x)-1)^2=(sqrt(3x+10))^2
=
3+x+1=3x+10

Get x on one side
X+4=3x+10
Subtract 4 from both sides
X=3x-7
Subtract 3x
-2x=-7
Divide by -2
X= 7/2

-
Square both sides. Collect terms. Square again. Collect terms. Use quadratic equation.

-
square both sides.
rearrange
solve
check for consistency,
i.e the terms under the root should be > 0
12
keywords: sqrt,solve,If,would,have,for,10,you,how,the,If you have the sqrt(3+x) -1 = sqrt(3x+10), how would you solve for x
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