Latest Huawei Aptitude Question SOLUTION: Let f(x) = ax2 + bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f (2) and that 3 is a root of f(x) = 0. 2. What is the other root of f(x)

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Let f(x) = ax2 + bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f (2) and that 3 is a root of f(x) = 0.

2. What is the other root of f(x) = 0?

Asked In Huawei chandreyee chowdhury (10 years ago)

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Ann Theressa  (10 years ago)

other root is -4
since 3 is a root and let y be the other root
(x-3)(x-y)=0
f(5)=2*(5-y)
f(2)=-1*(2-y)
f(5)=-3*f(2)
2*(5-y)=3*(2-y)
y=-4

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dayanand giri  (9 years ago)

A/C TO QUESTION
F(X)=0,AND 3 IS THE ROOT OF F(X)=0
THEN 9A+3b+C=0 .............. EQUATION FIRST
ALSO IT IS GIVEN THAT F(5)=-3F(2)
BY THIS EQUATION WE CAN GET THE EQUATION LIKE 25A+5B+C= -12A-6B-3C
IT WILL COME LIKE 37A=11B+4C=0 EQUATION SECOND
FOR SOLVING THESE TWO EQUATION WE CAN USE BRAZ GUNANKHAND METHOD
3 1 9 3 .................FROM EQUATION FIRST FROM B TO B COEFFICIENT VALUE
11 4 37 11 .................FROM EQUATION TWO " " " " " "
A/(3*4-11*1)= B/(1*37-4*9)= C/(9*11-37*3)
BY THIS THEOREM A=1,B=1,C= -12
PUT THESE VALUE IN F(X)=0
THE EXACT EQUATION WILL COME LIKE...... X^2+X-12=0
X^2+4X-3X-12=0
X(X+4)-3(X+4)=0
(X-3)(X+4)=0
X=3,-4
THEN OTHER ROOT MUST BE -4(NEGATIVE VALUE OF 4).....

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