Find the roots of the following equation: x + 1 x = 3 , x ≠ 0

Given, x + 1 x = 3 Multiplying throughout by x , we get x 2 + 1 = 3 x x 2 - 3 x + 1 = 0 , which is a quadratic equation. A quadratic equation a x 2 + b x + c = 0 has i two distinct real roots, if b 2 - 4 a c > 0 , i i two equal roots, if b 2 - 4 a c = 0 . i i i no real roots, if b 2 - 4 a c < 0 . Here, a = 1 , b = - 3 , c = 1 So, b 2 - 4 a c = 9 - 4 = 5 > 0 Therefore, x = 3 ± 5 2 So, the roots are 3 + 5 2 and 3 - 5 2

Find the roots of the equation: 1 x - 1 x - 2 = 3 , x ≠ 0 , 2

We have, 1 x - 1 x - 2 = 3 ⇒ x - 2 - x x x - 2 = 3 ⇒ - 2 x 2 - 2 x = 3 ⇒ 3 x 2 - 6 x = - 2 ⇒ 3 x 2 - 6 x + 2 = 0 We shall use quadratic formula to solve this equation. Comparing 3 x 2 - 6 x + 2 = 0 with a x 2 + b x + c = 0 , we get, a = 3 , b = - 6 , c = 2 . Now, we know, x = - b ± b 2 - 4 a c 2 a ⇒ x = - - 6 ± - 6 2 - 4 × 3 × 2 2 × 3 ⇒ x = 6 ± 36 - 24 6 ⇒ x = 6 ± 12 6 ⇒ x = 6 ± 2 3 6 ⇒ x = 3 ± 3 3 .

Find the discriminant of the equation 3 x 2 - 2 x + 1 3 = 0 and hence find the nature of its roots. Find them, if they are real.

Given equation: 3 x 2 - 2 x + 1 3 = 0 We know, for quadratic equation a x 2 + b x + c = 0 , the discriminant is b 2 - 4 a c . If b 2 - 4 a c > 0 , the equation has two distinct real roots. If b 2 - 4 a c = 0 , the equation has two equal roots. If b 2 - 4 a c < 0 , the equation has no real root. Now, discriminant for 3 x 2 - 2 x + 1 3 = 0 , - 2 2 - 4 × 3 × 1 3 = 4 - 4 = 0 As discriminant = 0 , we can say that the given equation have equal roots. The roots are - b 2 a , - b 2 a i.e 2 6 = 1 3 Thus, two roots are 1 3 , 1 3