# Which is the solution for the equation [( x + 7 )^1/2] + [(x-1)^1/2] = 4 ?

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### 3 Answers

[(x+7)^1/2]+[(x-1)^1/2=4 .....(1)

Let us multiply by (x+7)^1/2]-[(x-1)^1/2]

(x+7)-(x-1) = 4[(x+7)^1/2]-[(x-1)^1/2]

x+7-x+1= 4{[(x+7)^1/2]-[(x-1)^1/2]}\

8= 4{[(x+7)^1/2]-[(x-1)^1/2]}

Divide by 4:

2= [(x+7)^1/2]-[(x-1)^1/2] .....(1)

Let us add (1) and (2)

[(x+7)^1/2]+[(x-1)^1/2]+[(x+7)^1/2]-[(x-1)^1/2]=6

Reduce similar:

2(x+7)^1/2=6

Divide by 2:

(x+7)^1/2=3

x+7= 9

x= 2

To find the solution of (x+7)^(1/2)+(x-1)^(1/2) = 4.....(1)

Solution:

To rationalise the left we multiply by (x+7)^(1/2)-(x-1)^(1/2) both sides:

[(x+7)^(1/2) +(x-1)^(1/2)]*[(x+7)^(1/2) -(x-1)^(1/2)] = 4*[(x+7)^(1/2) -(x-1)^(1/2)] Or

(x+7)-(x-1) = 4[(x+7)^(1/2)-(x-1)^(1/2). Or

8=4(x+7)^(1/2)-(x-1)^(1/2). Or dividing by 4 both sides andswappig sides, we get:

(x+7)^(1/2)-(x-1)^(1/2) = 2....(2)

Eq(1) + Eq(2) gives: 2(x+7)^(1/2) = 4+2 = 6.

Or (x+7)^(1/2) = 3. Or

x+7 = 3^2. Or x = 3^2-7 Or

x = 2.

(1)-(2) gives: 2(x-1)^(1/2) = 4-2 = 2. Or

(x-1)^(1/2) = 1. Or

x-1 = 1^2 = 1 . Or

x = 1+1 = 2.

So , x = 2 is the solution:

Let's multiply the adjoint expression of the left side, to the both sides of the equation.

{[(x+7)^1/2] + [(x-1)^1/2]}X{[(x+7)^1/2] - [(x-1)^1/2]}= 4x{[(x+7)^1/2] - [(x-1)^1/2]}

Multiplying the paranthesis from the left side

[(x+7)^1/2]^2 - [(x+7)^1/2]x[(x-1)^1/2] + [(x+7)^1/2]x[(x-1)^1/2] - [(x-1)^1/2]^2= 4x{[(x+7)^1/2] - [(x-1)^1/2]}

(x+7) - (x-1)= 4x{[(x+7)^1/2] - [(x-1)^1/2]}

Opening the paranthesis from the left side

x + 7 -x +1= 4x{[(x+7)^1/2] - [(x-1)^1/2]}

8 = 4x{[(x+7)^1/2] - [(x-1)^1/2]}

o 2= {[(x+7)^1/2] - [(x-1)^1/2]}

Let's sum up this result with the initial equation

[(x+7)^1/2] - [(x-1)^1/2] + [(x+7)^1/2] + [(x-1)^1/2] = 6

One can note that two terms are the same: [(x+7)^1/2] and the other two term are opposite: [(x-1)^1/2], the last ones being reduced.

2x[(x+7)^1/2]= 6

[(x+7)^1/2]= 3

[(x+7)^1/2]^2= 3^2

x+7= 9

x= 9-7

x= 2

After verifying action, we can conclude that x= 2 is the solution of the equation.