Question

could anyone help me solve this problem please: sqt x-3 + sqt 2x-2 = 3

Answer 1

\[\sqrt{x-3} + \sqrt{2x-2} = 3\]\[\sqrt{2x-2} = 3 - \sqrt{x-3}\] Squaring both sides:
\[2x-2 = 9 - 6\sqrt{x-3} + x - 3\]\[6\sqrt{x-3} = 8 - x\]Again squaring both sides:
\[(6\sqrt{x-3})^{2}=(8-x)^{2}\]

Answer 2

@viande Welcome to Open Study.

Answer 3

thank you :) and btw is it necessary for the equations to be transposed?

Answer 4

not a compulsion.

Answer 5

If you don't transposed then you will get
\[2\sqrt{2x^{2}-8x+6}=8-3x\]