Answers

2016-04-28T16:13:02+05:30
Given :Sum of the zeroes 4
           one of the zero is 2+√5==α
another zero is 2-√5==β because 2+√5+β=4
                                    β=4-2+√5
                                    β=2-√5
So here in order to get the equation we need substitute in general formula of quadratic equation
x²-(α+β)+αb
x²-(4)x+(2+√5.2-√5)    {(a+b)(a-b)==a²-b²}
x²-4x+(4-5)
x²-4x+(-1)
x²-4x-1 is the answer
1 5 1
2016-04-28T16:33:11+05:30
Let,
the given zero = α = 2 +√5
let the other zero be = β
given ,
sum of zeroes = 4
α + β = 4
⇒ 2 +√5 +β = 4
⇒ β = 4-(2+√5)
⇒β = 2- √5
the quadratic polynomial whose zeroes are α and β is:
k[ x² -(α + β)x + αβ ] = 0
⇒ k[x² -{(2+√5)+(2-√5)}x + (2-√5)(2+√5) ] = 0
⇒ k [ x² - 4x +(4-5) ] = 0
⇒k[ x² - 4x -1]   = 0                   since (a² -b²) = (a+b)(a-b)
 if k = 1 the required equation is x² - 4x -1   = 0
 




0