Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^{2}$-6x+k, find the value of k such that $(\alpha- \beta )^{2}$ - 2$\alpha \beta$ = 40

Answer 1

Answer:

• k = -2 / 3

Step-by-step explanation:

Given that,

• α  and  β are zeroes of polynomial x² - 6 x + k
• ( α - β )² - 2 α β = 40

we need to find

• value of k

Comparing x² - 6 x + k with a x² + b x + c

we will get,

• a = 1
• b = - 6
• c = k

so,

→ sum of zeroes = α + β = -b/a = -(-6)/(1) = 6   ....eqn(1)

→ product of zeroes = α β = c/a = (k)/(1) = k   ....eqn(2)

Now,

as given

→ ( α - β )² - 2 α β = 40

[using algebraic identity ( a - b )² = a² + b² - 2 a b]

→ α² + β² - 2 α β - 2 α β = 40

→ ( α² + β² ) - 4 α β = 40

[using identity ( a + b )² - 2 a b = a² + b² ]

→ ( α + β )² - 2 α β - 4 α β = 40

→ ( α + β )² - 6 α β = 40

[using eqn(1) and (2)]

→ ( 6 )² - 6 ( k ) = 40

→ 36 - 6 k = 40

→ - 6 k = 4

→ k = 4 / -6 = -2 / 3

therefore,

value of k is -2 / 3 .