(a) ±2√2
(b) 2
(c) ±2
(d) √2
Doubt by Kanika
Solution :
2x²+kx-4=0
Here
a=2
Here
a=2
b=k
c=-4
We know,
D=b²-4ac
D=(k)²-4(2)(-4)
D=k²-(-32)
D=k²+32
D=b²-4ac
D=(k)²-4(2)(-4)
D=k²-(-32)
D=k²+32
Also, As per the quadratic formula
x=[-b±√D]/2a
'a' and 'b' are already rational numbers so in order to have rational roots, √D must be a rational number i.e. D must be a perfect square.
x=[-b±√D]/2a
'a' and 'b' are already rational numbers so in order to have rational roots, √D must be a rational number i.e. D must be a perfect square.
D=k²+32
If we put k=+1 [Why positive? Because positive value of k is being asked in the question]
D=(1)²+32=1+32=33 [Not a Perfect Square]
Similarly at K=2
D=(2)²+32=4+32=36 [A Perfect Square]
There could be many other possible values of k but we have been asked the least value of k.
Similarly at K=2
D=(2)²+32=4+32=36 [A Perfect Square]
There could be many other possible values of k but we have been asked the least value of k.
So, k=2
Hence, (b) 2 would be the correct option.