Question

Someone help me out please

Answer 1

1 step (B): raise both sides of the equation to the power of 2.

`(√(x+3)-√(2x-1))^2=(-2)^2,\\ (x+3)-2√(x+3)\cdot √(2x-1)+(2x-1)=4,\\ 3x+2-2√(x+3)\cdot √(2x-1)=4`.

2 step (A): simplify to obtain the final radical term on one side of the equation.

`-2√(x+3)\cdot √(2x-1)=4-3x-2,\\ -2√(x+3)\cdot √(2x-1)=2-3x,\\ 2√(x+3)\cdot √(2x-1)=3x-2`.

3 step (F): raise both sides of the equation to the power of 2 again.

`(2√(x+3)\cdot √(2x-1))^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2`.

4 step (E): simplify to get a quadratic equation.

`4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0`.

5 step (D): use the quadratic formula to find the values of x.

`D=(-32)^2-4\cdot 16=1024-64=960, \\ √(D) =8√(5) ,\\ x_(1,2)=(32\pm 8√(5))/(2) =16\pm 4√(5)`.

6 step (C): apply the zero product rule.

`x^2-32x+16=(x-16-4√(5)) (x-16+4√(5)) ,\\ (x-16-4√(5)) (x-16+4√(5)) =0,\\ x_1=16+4√(5) ,x_2=16-4√(5)`.

Additional 7 step: check these solutions, substituting into the initial equation.