p q r s be the 4 consicutive integers from 1 to 12 what is the least sum for(p+q)+(r/s)
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Step-by-step explanation:
x+p
1
+
x+q
1
=
r
1
(x+p)(x+q)
x+p+x+q
=
r
1
(2x+p+q)r=x
2
+px+qx+pq
x
2
+(p+q−2r)x+pq−pr−qr=0
Let α and β be the roots.
⟹α+β=−(p+q−2r) ...[1]
⟹αβ=pq−pr−qr ...[2]
Roots are equal in magnitude and opposite in sign
⟹α+β=0 .
⟹−(p+q−2r)=0 ...[3]
α
2
+β
2
=(α+β)
2
−2αβ
=(−(p+q−2r))
2
−2(pq−pr−qr) ...(from [1] and [2])
=p
2
+q
2
+4r
2
+2pq−4pr−4qr−2pq+2pr+2qr
=p
2
+q
2
+4r
2
−2pr−2qr
=p
2
+q
2
+2r(2r−p−q) ...(from [3])
=p
2
+q
2
+0
=p
2
+q
2
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