If #3x^2+2(p+q+r)x+pq+qr+rp# be a perfect squares,prove that #p=q=r#?

#3x^2+2(p+q+r)x+pq+qr+rp#

1 Answer
George C.
Jan 18, 2018

See explanation...

Explanation:

#3(3x^2+2(p+q+r)x+pq+qr+rp)#

#=9x^2+6(p+q+r)x+3(pq+qr+rp)#

#=(3x)^2+2(3x)(p+q+r)+(p+q+r)^2+3(pq+qr+rp)-(p+q+r)^2#

#=(3x+(p+q+r))^2+3(pq+qr+rp)-(p+q+r)^2#

If this is a square for all values of #x# then:

#3(pq+qr+rp)-(p+q+r)^2 = 0#

So:

#0 = -2(3(pq+qr+rp)-(p+q+r)^2)#

#color(white)(0) = 2p^2+2q^2+2r^2-2pq-2qr-2rp#

#color(white)(0) = p^2-2pq+q^2+q^2-2qr+r^2+r^2-2rp+p^2#

#color(white)(0) = (p-q)^2+(q-r)^2+(r-p)^2#

This is only possible for real values of #p, q, r# if:

#p = q = r#

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