If p and q are the zeroes of polynomial f(x)=2x^2-7x+3 find p^2+q^2
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Answer is 37/4 of this question....
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p+q=7/2 pq=3/2
(p+q)^2=(p-q)^2+4pq
49/4=(p-q)^2+4×3/2
(p-q)^2=49/4-6
(p-q)^2=25/4
p-q=5/2 or -5/2
p+q=7/2
p-q=5/2
2p=12/2
p=3
q=7/2-p
=7/2-3
=1/2
Now p^2 +q^2=9+1/4=37/4 And
Second solution (short)
(p+q)^2=p^2+q^2+2pq
49/4=p^2+q^2+2×3/2
p^2 +q^2=49/4-3
=37/4 Ans
(p+q)^2=(p-q)^2+4pq
49/4=(p-q)^2+4×3/2
(p-q)^2=49/4-6
(p-q)^2=25/4
p-q=5/2 or -5/2
p+q=7/2
p-q=5/2
2p=12/2
p=3
q=7/2-p
=7/2-3
=1/2
Now p^2 +q^2=9+1/4=37/4 And
Second solution (short)
(p+q)^2=p^2+q^2+2pq
49/4=p^2+q^2+2×3/2
p^2 +q^2=49/4-3
=37/4 Ans
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