Question

Arrange the steps to solve the equation

`√(x + 3) - √(2x - 1) = - 2`

Answer 1

`Domain:\\\\x+3\geq0\ \wedge\ 2x-1\geq0\\\\x\geq-3\ \wedge\ x\geq0.5\\\\D:x\in[0.5;\ \infty)`

`√(x+3)-√(2x-1)=-2\qquad\text{square of both sides}\\\\(√(x+3)-√(2x-1))^2=(-2)^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(√(x+3))^2-2(√(x+3))(√(2x-1))+(√(2x-1))^2=4\qquad\text{use}\ (√(a))^2=a\\\\x+3-2√((x+3)(2x-1))+2x-1=4\qquad\text{combine like terms}\\\\3x+2-2√((x+3)(2x-1))=4\qquad\text{subtract 3x + 2 from both sides}\\\\-2√((x+3)(2x-1))=2-3x\qquad\text{square of both sides}\\\\\left(-2√((x+3)(2x-1))\right)^2=(2-3x)^2`

`4(x+3)(2x-1)=2^2-2(2)(3x)+(3x)^2\qquad\text{use distributive property}\\\\(4x+12)(2x-1)=4-12x+9x^2\\\\(4x)(2x)+(4x)(-1)+(12)(2x)+(12)(-1)=4-12x+9x^2\\\\8x^2-4x+24x-12=4-12x+9x^2\\\\8x^2+20x-12=4-12x+9x^2\qquad\text{subtract }\ 9x^2\ \text{from both sides}\\\\-x^2+20x-12=4-12x\qquad\text{add 12 and}\ 12x\ \text{from both sides}\\\\-x^2+36x=16\qquad\text{change the signs}\\\\x^2-36x=-16\\\\x^2-2(x)(16)=-16\qquad\text{add}\ 16^2\ \text{to both sides}`

`\underbrace{x^2-2(x)(16)+16^2}_((a-b)^2=a^2-2ab+b^2)=16^2-16\\\\(x-16)^2=256-16\\\\(x-16)^2=240\iff x-16=\pm√(240)\\\\x-16=\pm√(16\cdot15)\\\\x-16=-√(16)\cdot√(15)\ \vee\ x-16=√(16)\cdot√(15)\\\\x-16=-4√(15)\ \vee\ x-16=4√(15)\qquad\text{add 16 to both sides}\\\\x=16-4√(15)\\otin D\ \vee\ x=16+4√(15)\\\\\boxed{x=4(4+√(15))}`