Find the nature of the roots of the quadratic equations. If the real roots exist, find them; 2x2 – 6x + 3 = 0

Asked by Abhisek | 1 year ago |  166

Solution :-

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.

$$x= \dfrac{-b±\sqrt{b^2-4ac}}{2a}$$

$$x= \dfrac{(-6)±\sqrt{(-6)^2-4(2)(3)}}{2(2)}$$

$$x= \dfrac{6±\sqrt{12}}{4}=\dfrac{6±2\sqrt{3}}{4}$$

$$x= \dfrac{3±\sqrt{3}}{2}$$

Therefore the roots for the given equation are $$\dfrac{3+\sqrt{3}}{2} \;and\; \dfrac{3-\sqrt{3}}{2}$$

Answered by Pragya Singh | 1 year ago

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