Solve. √2x+6-√x+4=1

Question

Okay the 2x+6 is all under the radical and the x+4 is also under the radical please help

Answer 1

Sqrt(2x+6)-sqrt(x+4) = 1

Squring to each side.

[Sqrt(2x+6)-sqrt(x+4)]^2 = 1^2

[Sqrt(2x+6)]^2+[Sqrt(x+4)]^2-2[Sqrt(2x+6)sqrt(x+4) = 1

2x+6 + x+4 -2Sqrt[(2x+6)(x+4)]= 1

3x+10-2sqrt[2x^2+8x+6x+24] = 1

3x+10-2sqrt[2x^2+14x+24] = 1

Subract 3x,10 from each side.

3x+10-3x-10-2sqrt[2x^2+14x+24] = 1-3x-10

-2sqrt[2x^2+14x+24] = -3x-9

Multple to each side by negitive one.

2sqrt[2x^2+14x+24] = 3x+9

Again sqauring to each side.

4(2x^2+14x+24) = 9x^2+81+54x

8x^2+56x+96 = 9x^2+81+54x

Bring all terms to one side.

9x^2+81+54x-8x^2-56x-96  = 0

x^2-2x-15 = 0

x^2-5x+3x-15 = 0

x(x-5)+3(x-5) = 0

(x-5)+(x+3) = 0

x-5 = 0 and x+3 = 0

x-5+5 = 0+5 and x+3-3 = 0-3

x = 5 and x = -3

Comments

The radical inequality

Check your answers to avoid extraneous roots:

For x = 5

Above statement is true.

For x = -3

Above statement is false.

Solution of is x = 5.