Question

I just need the extraneous root

Answer 1

Answer:

Step-by-step explanation:

$2+\sqrt{2x-3} =\sqrt{x+7} ~~~(1)\\ x \ge -7\\and ~x\ge \frac{3}{2} \\squaring ~(1)\\4+2x-3+4\sqrt{2x-3} =x+7\\4\sqrt{2x-3} =x+7-2x-4+3\\4\sqrt{2x-1} =-x+6\\squaring\\16(2x-3)=x^2-12x+36\\32x-48=x^2-12x+36\\x^2-12x-32x+36+48=0\\x^2-44x+84=0\\\\x^2-2x-42x+84=0\\x(x-2)-42(x-2)=0\\(x-2)(x-42)=0\\x=2,42\\x=2 is a solution.\\whereas ~x=42 ~is~an~extraneous ~root.\\as~2+\sqrt{2*42-3} =\sqrt{42+7} \\2+\sqrt{81} =\sqrt{49} \\2+9=7\\or 11=7\\which~is~not~true.$