Answer to Question #302289 in Algebra for Carmen

Question #302289

If the equation (1-q+(p^2)/2)x^2+p(1+q)x+q(q-1) +((p^2)/2)=0 has two equal roots, prove that p^2=4q


1
Expert's answer
2022-02-28T09:05:05-0500

Quadratic equation has two equal roots if


1q+p2201-q+\dfrac{p^2}{2}\not=0

D=(p(1+q))24(1q+p22)(q(q1)+p22)=0D=(p(1+q))^2-4(1-q+\dfrac{p^2}{2})(q(q-1)+\dfrac{p^2}{2})=0

p2(1+q)2(22q+p2)(2q22q+p2)=0p^2(1+q)^2-(2-2q+p^2)(2q^2-2q+p^2)=0

p2(1+2q+q2)(22q)(2q22q)p2(22q)p^2(1+2q+q^2)-(2-2q)(2q^2-2q)-p^2(2-2q)

p2(2q22q+p2)=0-p^2(2q^2-2q+p^2)=0

p4+p2(1+2q+q22+2q2q2+2q)-p^4+p^2(1+2q+q^2-2+2q-2q^2+2q)

(22q)(2q22q)=0-(2-2q)(2q^2-2q)=0

p4+p2(q26q+1)4q(q1)2=0p^4+p^2(q^2-6q+1)-4q(q-1)^2=0

If p2=4qp^2=4q


16q2+4q324q2+4q4q3+8q24q=0,16q^2+4q^3-24q^2+4q-4q^3+8q^2-4q=0,

True for qR.q\in \R.

Or


p2=4q(q1)24q=(q1)2p^2=\dfrac{-4q(q-1)^2}{4q}=-(q-1)^2

The only solution is p2=0p^2=0 for q=1.q=1.

But in this case


1q+p22=11+0=01-q+\dfrac{p^2}{2}=1-1+0=0

and we have an equation


0=00=0

Hence the quadratic equation


p4+p2(q26q+1)4q(q1)2=0p^4+p^2(q^2-6q+1)-4q(q-1)^2=0

has the only solution which satisfies our task


p2=4q.p^2=4q.

In this case the quadratic equation


(1q+p22)x2+p(1+q)x+q(q1)+p22=0(1-q+\dfrac{p^2}{2})x^2+p(1+q)x+q(q-1)+\dfrac{p^2}{2}=0

has two equal roots.


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