JAMB - Mathematics (1986 - No. 21)
Solve the equation 3x\(^2\) + 6x - 2 = 0
x = -1 \(\pm\) \(\frac{\sqrt{3}}{3}\)
x = -1 \(\pm\) \(\frac{\sqrt{15}}{3}\)
x = -2 \(\pm\) 2
x = 3 \(\pm\) \(\frac{\sqrt{3}}{15}\)
Explanation
3x2 + 6x - 2 = 0
Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\)
a = 3, b = 6, c = -2
x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\)
x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\)
x = \(\frac{-6 \pm \sqrt{60}}{6}\)
x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\)
x = \(-1 \pm \frac{\sqrt{15}}{3}\)
Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\)
a = 3, b = 6, c = -2
x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\)
x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\)
x = \(\frac{-6 \pm \sqrt{60}}{6}\)
x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\)
x = \(-1 \pm \frac{\sqrt{15}}{3}\)
Comments (0)
