**(i) x**

^{2}+ 3x – 4 = 0, x = 1, –2, – 3

**Sol.**

**a) By putting x = 1 in L.H.S. we get**

L.H.S. = (1)

**+ 3(1) – 4**^{2}
= 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)

**+ 3(–2) – 4**^{2}
= 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)

**+ 3(–3) – 4**^{2}
= 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.**