Oxford Mathematics For The New Century 2a Answer
( x = 5, y = 4 ) (correct result, but no working shown).
Algebraically: ( 2x^2 - 5x - 3 = 0 ) → divide by 2: ( x^2 - \frac52x - \frac32 = 0 ). Add ( \left(\frac54\right)^2 = \frac2516 ) to both sides: ( x^2 - \frac52x + \frac2516 = \frac32 + \frac2516 = \frac2416 + \frac2516 = \frac4916 ). LHS: ( (x - \frac54)^2 = \frac4916 ) → ( x - \frac54 = \pm \frac74 ) → ( x = 3 ) or ( x = -\frac12 ).
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Navigating Oxford Mathematics for the New Century 2A: A Guide to Solutions and Success LHS: ( (x - \frac54)^2 = \frac4916 )
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