x2 + 5x – 14 = 0, x = √2 , –7, 3

(iii) x² + 5x – 14 = 0, x = √2 , –7, 3

Sol. a) By putting x =  √2 in L.H.S. we get

L.H.S. = ( √2 )² + 5( √2 ) – 14

= 2 + 5√ 2 – 14

= 5 √2 – 12

≠ R.H.S.

∴  L.H.S. ≠ R.H.S.

Thus equation is not satisfied.

So,  √2 is not the root of the given quadratic equation.

b) By putting x = –7 in L.H.S. we get

L.H.S. = (–7)² + 5(–7) – 14

= 49 – 35 – 14

= 49 – 49

= 0

= R.H.S.

∴ L.H.S. = R.H.S.

Thus equation is satisfied.

So,  –7 is the root of the given quadratic equation.

c) By putting x = 3 in L.H.S. we get

L.H.S. = (3)² + 5(3) – 14

= 9 + 15 – 14

= 24 – 14

= 10

≠ R.H.S.

∴ L.H.S. ≠ R.H.S.

Thus equation is not satisfied.

So,  3 is not the root of the given quadratic equation.