Quadratic equation has two equal roots if
1−q+2p2=0
D=(p(1+q))2−4(1−q+2p2)(q(q−1)+2p2)=0
p2(1+q)2−(2−2q+p2)(2q2−2q+p2)=0
p2(1+2q+q2)−(2−2q)(2q2−2q)−p2(2−2q)
−p2(2q2−2q+p2)=0
−p4+p2(1+2q+q2−2+2q−2q2+2q)
−(2−2q)(2q2−2q)=0
p4+p2(q2−6q+1)−4q(q−1)2=0 If p2=4q
16q2+4q3−24q2+4q−4q3+8q2−4q=0, True for q∈R.
Or
p2=4q−4q(q−1)2=−(q−1)2The only solution is p2=0 for q=1.
But in this case
1−q+2p2=1−1+0=0and we have an equation
0=0 Hence the quadratic equation
p4+p2(q2−6q+1)−4q(q−1)2=0 has the only solution which satisfies our task
p2=4q. In this case the quadratic equation
(1−q+2p2)x2+p(1+q)x+q(q−1)+2p2=0has two equal roots.
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