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## Answers

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

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**