Find the nature of the roots of the quadratic equations. If the real roots exist, find them; 3x2 – $$4 \sqrt{3}x$$ + 4 = 0

Asked by Abhisek | 1 year ago |  124

##### Solution :-

3x2 – $$4 \sqrt{3}x$$ + 4 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 3, b = $$-4\sqrt{3}$$ and c = 4

We know, Discriminant = b2 – 4ac

= $$( -4\sqrt{3})^2$$ – 4(3)(4)

= 48 – 48 = 0

As b2 – 4ac = 0,

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be $$\dfrac{-b}{2a}\;and\;\dfrac{-b}{2a}$$

$$\dfrac{-b}{2a}=\dfrac{(-4\sqrt{3})}{2\times 3}$$

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

Therefore, the roots are $$\dfrac{2}{\sqrt{3}}\;and\;\dfrac{2}{\sqrt{3}}$$

Answered by Pragya Singh | 1 year ago

### Related Questions

#### A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

#### Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

#### The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $$\dfrac{29}{20}$$ Find the original fraction.