(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.
