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Grade: 11

                        

Q.15 nd Q.16 from the attached image. Thanks in advance

2 years ago

Answers : (2)

Aditya Gupta
2065 Points
							
let roots be a,b,x1 and a,b,x2
now, a+b+x1= – 5
and  a+b+x2= – 7
subtract
x1 – x2=2
now, ab+ax1+bx1=p
and ab+ax2+bx2=p
suntract
a+b=0 put this in above equation
a+b+x2= – 7 oe x2= – 7 and x1= – 5
so, (-5,-7) is correct
the 16 th question is quite a nice question.
ket f(x)=ax^2+bx+c
then f(1)=a+b+c0, that would imply the existence of a real root. hence,  f(x)
now, f(-2)=4a-2b+c
or 4a+c
 
2 years ago
Aditya Gupta
2065 Points
							
there is something terribly wrong with this site. i had typed a complete answer above, but due to technical error it doesnt show.
f(1) is less than zero.
hence f(-2) is less than zero
so option b is correct
 
2 years ago
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