After eliminating radicals, what quadratic equation can you solve to find the potential solutions of sqrt 2x+3 - sqrt x+1 = 1

Step-by-step explanation:

We have given:

√2x+3 - √x+1  = 1

First of all isolate the square root of the left hand side:

√2x+3 = √x+1 +1

Now take square on both sides.

(√2x+3)^2 = (√x+1 +1)^2

Open the R.H.S by squaring formula.

∴(a+b)^2 = a^2+2ab+b^2

2x+3 = (√x+1)^2 + 2(√x+1)(1)+(1)^2

2x+3 = x+1 +2√x+1 +1

2x+3 = x+2 +2√x+1

Combine the like terms:

2x-x+3-2 = 2√x+1

x+1 = 2√x+1

Take square on both sides

(x+1)^2 = (2√x+1)^2

x²+2x+1 = 4x+4

x²+2x-4x+1-4 = 0

x²-2x-3 = 0

Now solve the quadratic equation:

a = 1 , b= -2 , c = -3

x = -b+/-√b²-4ac/2a

x = -(-2)+/-√(-2)² - 4(1)(-3) / 2(1)

x = 2 +/- √4+12 / 2

x = 2+/- √16/2

x = 2+/- 4 /2

x = 2+4/2     , x = 2-4/2

x = 6/2      , x = -2/2

x = 3      , x = -1

The solutions we get is (3, -1).

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