WAEC SSCE - Further Mathematics - 2010…

The roots of a quadratic equation are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$. Find its equation.

$x^{2} - 6x - 9 = 0$

$x^{2} - 6x + 6 = 0$ Respuesta correcta

$x^{2} + 6x - 9 = 0$

$x^{2} + 6x + 6 = 0$

Detalles de la respuesta

The roots of a quadratic equation of the form $ax^{2} + bx + c = 0$ are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$, then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$ is $x^{2} - 6x + 6 = 0$. Therefore, the answer is.






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WAEC SSCE - Further Mathematics - 2010 (Objective)