Finding Constants for Quadratic Equations

[anemone]

Find constants $P,\,Q,\,R,\,S, a,\,b$ such that

$P(x-a)^2+Q(x-b)^2=5x^2+8x+14$

$R(x-a)^2+S(x-b)^2=x^2+10x+7$

 

[Albert1]

Find constants $P,\,Q,\,R,\,S, a,\,b$ such that

$P(x-a)^2+Q(x-b)^2=5x^2+8x+14$

$R(x-a)^2+S(x-b)^2=x^2+10x+7$

Spoiler

compare both sides we get:
$P+Q=5---(1)$
$aP+bQ=-4---(2)$
$a^2P+b^2Q=14---(3)$
$R+S=1---(4)$
$aR+bS=-5---(5)$
$a^2R+b^2S=7---(6)$
from (1)(2)(3) we get :
$-b=\dfrac {14+4a}{5a+4}---(7)$
from (4)(5)(6) we get :
$-b=\dfrac {7+5a}{a+5}--(8)$
from (7)(8) we get :
$a=-2,1$
the rest is easy ,and the solutions will be:
$(a,b,P,Q,R,S)=(-2,1,3,2,2,-1)$
or:
$(a,b,P,Q,R,S)=(1,-2,2,3-1,2)$

 

[anemone]

Thanks for participating, Albert and your answer is correct!:)