Question

find the quadratic polynomial whose zeros are root 3 + 5 and root 3 minus 5​

Answer 1

Answer:

x² - 2√3 - 22

Step-by-step explanation:

It is given that √3 + 5 and √3 - 5 are the zeroes of the required polynomial.

Let the two zeroes be α and β of the required polynomial.

∴ α = √3 + 5, β = √3 - 5

_____________________________

Now,

• Sum of zeroes = α + β

→ (√3 + 5) + (√3 - 5)

→ √3 + 5 + √3 - 5

→ √3 + √3

→ 2√3

• Product of zeroes = αβ

→ (√3 + 5)(√3 - 5)

• Identity : (a - b)(a + b) = a² - b²

Here, a = √3, b = 5

→ (√3)² - (5)²

→ 3 - 25

→ - 22

_____________________________

The required polynomial is :

→ p(x) = k [ x² - (α + β)x + αβ ]

• Putting known values.

→ p(x) = k [ x² - (2√3)x + (- 22) ]

→ p(x) = k [x² - 2√3x - 22]

• Putting k = 1.

→ p(x) = x² - 2√3 - 22

Answer 2

☀ QUESTION :- ☀

find the quadratic polynomial whose zeros are √ 3 + 5 and √3 - 5.

☀ ANSWER :- ☀

➠ let the two zeroes be α & β .

➠ .°. α = √3 + 5 & β = √3 -5

➠ sum of zeroes = α + β

➠ (√3 +5) + (√3 - 5)

➠ √3 + 5 + √3 -5

➠ √3 + √3

➠ 2√3.

➠ sum of product = α × β

➠ ( √3 +5 ) × (√3 -5)

➠ (√3)² - (5)² [(°.° (a +b) (a-b) = a² - b²)]

➠ 3 - 25

➠ -22.

WE KNOW THAT,

➠ p(x) = k(x²-(α+β)x + α×β)

NOW,

➠ p(x) = K ( x² - ( 2√3)x +(-22)

➠ p(x) = K (x ² - 2√3x -22)

➠ p(x) = (x² - 2√3x -22) [( °.° put k = 1)]

SO ,FINAL ANSWER IS.

➠ p(x) = x² - 2√3x - 22.