Show that the equation x²+ax-4=0 has distinct real roots
AbaAaliya:
plz answer it fast guys
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A=1,B=a,C=-4
D=B^2- 4AC
D=(a)^2-4(1)(-4)
D=a^2+16
since square of no. or variable cannot be negative hence D>0
D=B^2- 4AC
D=(a)^2-4(1)(-4)
D=a^2+16
since square of no. or variable cannot be negative hence D>0
But it is not negative... No negative sign is there in your ans.
negative in the sense the value of D is more 0
i.e.D>0
D>0
Oh yeah... Sorry
it's ok
...
plz mark it as brainliest ans
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