(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.
