Question

If a,b,c and d are natural numbers such that a square +b square=41 and c square +d square =25 then find the polynomial whoes zeros are (a+b) and (c+d).

Answer 1

★ QUADRATICS RESOLUTION ★

☣ GIVEN THAT a,b,c,and d ARE REAL NUMBERS ...

☣ CAN'T SAY ABOUT BEING SAME OR DISTINCTIVE ...

☣ HENCE PROCEEDING IN THE MANNER ...

☣ a² + b² = 41

☣ c² + d² = 25

☣ AND THIS REAL SET OF EQUALITY IS SATISFIED BY ONLY POSSIBLE NEAR SQUARES ASLIKE ...

☣ LET , a = 4 , b= 5 , c= 3 , d = 4

☣ ACCORDING TO QUESTION ...
THE POSSIBILITIES OR THE POSSIBLE COMBINATIONS OF THIS EQUATIONS CAN BE THIS VALUE ONLY ...
OTHERWISE , NO OTHER VALUES OF REAL NUMBERS SATISFY THIS ...

☣ NOW , POLYNOMIAL IN "X" HAVING ROOTS AS (a+b) AND (c + d) IS ...

☣ X² - [ ( a+b) + (c+d) ] X + (a +b)(c+d) =0

☣ X² - [ 16 ] X + ( 9)(7) =0

☣ X² - 16X + 63 =0

∴ IT'S THE REQUIRED POLYNOMIAL WHICH SATISFIES THE GIVEN CONDITIONS IN EVERY POSSIBLE WAY ...

☣ AS THE QUESTION IS ASKING OF POSSIBILITY ...

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