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Quadratic Equations
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30/05/2021 12:03 pm
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Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
2x2 – 6x + 3 = 0
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30/05/2021 12:10 pm
2x2 – 6x + 3 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
As we know, Discriminant = b2 – 4ac
= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac > 0,
Therefore, there are distinct real roots exist for this equation, 2x2 – 6x + 3 = 0.
x = \(\frac{-(-6) \pm \sqrt{-6^2 - 4(2)(3)}}{2(2)}\)
= \(\frac{6 \pm 2 \sqrt{3}}{4}\)
= \(\frac{3 \pm \sqrt{3}}{4}\)
= \(\frac{3 \pm \sqrt{3}}{2}\)
Therefore the roots for the given equation are \(\frac{3 + \sqrt{3}}{2}\) and \(\frac{3 - \sqrt{3}}{2}\)