(i) x2 + 3x – 4 = 0, x = 1, –2, – 3
Sol. a) By putting x = 1 in L.H.S. we get
L.H.S. = (1)2 + 3(1) – 4
= 1 + 3 – 4
= 4 – 4
= 0
= R.H.S.
∴ L.H.S. = R.H.S.
Thus equation is satisfied.
So, 1 is the root of the given quadratic equation.
b) By putting x = –2 in L.H.S. we get
L.H.S. = (–2)2 + 3(–2) – 4
= 4 – 6 – 4
= – 6
≠ 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.
c) By putting x = –3 in L.H.S. we get
L.H.S. = (–3)2 + 3(–3) – 4
= 9 – 9 – 4
= – 4
≠ 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.
