A tricky Admission Algebra Problem from Oxford University Interview

[@YAWTon]

Square both sides of the equation: x+2√3√x+3=2x-7, therefore x-2√3√x-10=0. Solve for √x using formula for quadratic equation: √x=√3±√13. Discard √x=√3-√13

[@9허공]

For real solution, we first define the domain of x. √x ≥ 0 and √(2x - 7) ≥ 0 and √(2x - 7) > √3 => x > 8. So x =16 - 2√39 is discarded immediately.

[@RealQinnMalloryu4]

{x+x ➖ }+{3+3 ➖ }={x^2+6}=,6x^2 3^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).(2x)^2(7)^2={4x^2 ➖ 49}=45x^2 3^15x^2 3^3^5x^2 1^3^1x^2 3x^2 (x ➖ 3x+2).

[@chukkareddyakkenapally5487]

Your teaching is boring

[@superacademy247]

Why?