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Q.15 nd Q.16 from the attached image. Thanks in advance
let roots be a,b,x1 and a,b,x2now, a+b+x1= – 5and a+b+x2= – 7subtractx1 – x2=2now, ab+ax1+bx1=pand ab+ax2+bx2=psuntracta+b=0 put this in above equationa+b+x2= – 7 oe x2= – 7 and x1= – 5so, (-5,-7) is correctthe 16 th question is quite a nice question.ket f(x)=ax^2+bx+cthen f(1)=a+b+c0, that would imply the existence of a real root. hence, f(x)now, f(-2)=4a-2b+cor 4a+c
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 zeroso option b is correct
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