2p2 + 5p – 3 = 0, p = 1, ½, –3


(iv) 2p2 + 5p – 3 = 0, p = 1, ½,  –3



Sol. a) By putting p = 1 in L.H.S. we get

L.H.S. = 2(1)2 + 5(1) – 3

= 2(1) + 5 – 3

= 2 + 5 – 3

= 7 – 3

= 4

≠ R.H.S.

L.H.S. ≠ R.H.S.

Thus equation is not satisfied.

So,  1 is not the root of the given quadratic equation.



b) By putting p = ½ in L.H.S. we get

L.H.S. = 2( ½)2 +5( ½ ) – 3

= 2( ¼ )  + 5/2 – 3

= ½ + 5/2 – 3

= (1+5)/2 – 3

= 6/2 – 3

= 3 – 3

= 0

= R.H.S.

L.H.S. = R.H.S.

Thus equation is satisfied.

So, ½, is the root of the given quadratic equation.



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

L.H.S. = 2(– 3)2 + 5(– 3) – 3

= 2(9) – 15 – 3

= 18 – 18

= 0

= R.H.S.

L.H.S. = R.H.S.

Thus equation is satisfied.
So,  – 3 is the root of the given quadratic equation.