Question

If a, ß are zeroes of P(x) = x² - 6x + k, such that a² + ² = 20 then find the value of k.​

Answer 1

$\sf{\huge{\underline{\green{Given :-}}}}$

• a, ß are zeroes of P(x) = x² - 6x + k, such that α² + β² = 20 .

$\sf{\huge{\underline{\green{To\:Find :-}}}}$

• The value of k.

$\sf{\huge{\underline{\green{Answer :-}}}}$

We have,

f(x) = x2 - 6x + k

α² + β² = 20

➝ ( α + β )² - 2αβ = 20 ------(1)

α + β = -b/a

➝ -(-6)/1

➝ 6

αβ = c/a

➝ k/1

➝ k

Putting Value in (1),

( α + β )² - 2αβ = 20

➝ ( 6 )² - 2k = 20

➝ 36 - 2k = 20

➝ - 2k = 20 - 36

➝ - 2k = -16

➝ k = -16/-2

➝ k = 8

The value of k is 8.