Question Number 215072 by hardmath last updated on 27/Dec/24
![x^2 + 2000x + 1 = 0 Roots: a and b x^2 − 2008x − 1 = 0 Roots: c and d Find: (a+c)(b+d)(a−d)(b−c) = ?](https://www.tinkutara.com/question/Q215072.png)
Commented by TonyCWX08 last updated on 28/Dec/24
![Next time. Have your answer as ANSWER. Not COMMENT.](https://www.tinkutara.com/question/Q215085.png)
Commented by Abdullahrussell last updated on 28/Dec/24
![(a+c)(b+d)(a−d)(b−c) =(ab+ad+bc+cd)(ab−ac−bd+cd) =(1+ad+bc−1)(1−ac−bd−1) =−(ad+bc)(ac+bd) =−(a^2 cd+abd^2 +abc^2 +b^2 cd) =−(−a^2 +d^2 +c^2 −b^2 ) =a^2 +b^2 −(c^2 +d^2 ) =(a+b)^2 −2ab−(c+d)^2 +2cd =(−2000)^2 −2−(2008)^2 −2 =(2000+2008)(2000−2008)−4 =4008×−8−4 =−32064−4=−32068](https://www.tinkutara.com/question/Q215076.png)
Commented by MathematicalUser2357 last updated on 28/Dec/24
You better not to be angry
Commented by mr W last updated on 29/Dec/24
![i think he is not angry. he is just giving an useful suggestion.](https://www.tinkutara.com/question/Q215136.png)
Answered by TonyCWX08 last updated on 28/Dec/24
![a+b=−2000 ab=1 c+d=2008 cd=−1 (a+c)(b+d) =ab+ad+bc+cd =ad+bc (a−d)(b−c) =ab−ac−bd+cd =−ac−bd (ad+bc)(−ac−bd) =−a^2 cd−abd^2 −ac^2 b−b^2 cd =a^2 −d^2 −c^2 +b^2 =a^2 +b^2 −(c^2 +d^2 ) =((a+b)^2 −2ac)−((c+d)^2 −2cd) =2000^2 −2−(2008^2 +2) =2000^2 −2008^2 −4 =−32068](https://www.tinkutara.com/question/Q215077.png)