Question

if α,β are the zeros of the polynomial p(x)=2x^2+5x+p,satisfying the relationship α^2+β^2+α+β=21upon4, then find the value of p
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Answer 1

here is your ans.............

Answer 2

α+β=-b/a α×β=c/a

=-5/2 =p/2

α^2+β^2+α+β=21/4

(α+β)^2-2αβ+α+β=21/4

(-5/2)^2-2(p/2)+(-5/2)=21/4

25/4-p-5/2=21/4

25/4-21/4=p+5/2

(25-21)/4=(2p+5)/2

4/4=(2p+5)/2

1=(2p+5)/2

1×2=2p+5

2=2p+5

2-5=2p

-3=2p

2p=-3

p=-3/2

THE VALUE OF P IS -3/2

《《《《《 VARIFICATION》》》》》

α^2+β^2+α+β=21/4

(α+β)^2-2αβ+α+β=21/4

(-5/2)^2-2(-3/2/2)+(-5/2)=21/4

25/4-2(-3/2×1/2)-5/2=21/4

25/4-2(-3/4)-5/2=21/4

25/4-1(-3/2)-5/2=21/4

25/4+3/2-5/2=21/4

25/4+3/2×2/2-5/2×2/2=21/4

25/4+6/4-10/4=21/4

(25+6-10)/4=21/4

(31-10)/4=21/4

21/4=21/4

LHS=RHS

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