This question was asked in Ex 4.2, 1

Jaswant Singh

- Jan. 22, 2017, 4:53 p.m.Teachoo Expert
Hi Jaswant,

The text version of your answer

Find the roots of the quadratic equation √3

*x*2 – 2√2 x – 2√3 = 0√3

*x*2 – 2√2 x – 2√3 = 0Comparing equation with ax2 + bx + c = 0

Here, a = √3, b = – 2√2 , c = –2√3

We know that

D = b2 – 4ac

= ("– 2" √2)^2– 4 ×(√3)×(-2√3)

= (4 ×2)+8(√3 ×√3)

= 8 + 8 × 3

= 8 + 24

= 32

So, the roots of the equation is given by

x = (- b ± √D)/2a

*Putting values*x = (- (- 2√2) ± √32)/(2 × √3)

x = (2√2 ± √(16 × 2))/(2 × √3)

x = (2√2 ± √16 × √2)/(2√3)

x = (2√2 ± 4 √2)/(2√3)

x = (2√2 + 4√2)/(2√3)

x = (6√2)/(2√3) = (3√2)/√3 = √3 × √2

x = √(3 ×2) =

**√6**x = (2√2 - 4 √2)/(2√3)

x = (-2√2)/(2√3)

x = (-√2)/√3 =

**-√(2/3)**Hence, the roots of the equation are √6 and -√(2/3).

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