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## Answers

[(x-2)/(x-1)]×[(x+4)/(2x+2)] = 0

[(x-2)(x+4)] / [(x-1)(2x+2)] = 0

(x² + 4x - 2x - 8)/ (2x² + 2x - 2x -2) = 0

(x² + 2x - 8) / (2x² - 2) = 0

x² + 2x - 8 = 0 × (2x² - 2)

⇒x² + 2x - 8 = 0 ------------------- eq (1)

Standard eq of Q.E is ax² + bx +c = 0 --------- (2)

Comparing (1) and (2) we get

a = 1 , b = 2 and c= -8

With the help of discriminant we can find the nature of their roots.

Discriminant(Δ) = b² - 4ac

2² - 4(1)(-8) = 4+32 = 36

If b² - 4ac is greater than 0 then the roots are

**REAL AND DISTINCT**

∴The nature of the roots are

**real and distinct.**