Question

If α &β are the zeros of the polynomial f(x)=x²-6x+k, find the value of k such that α²+β²=40

Answer 1

Hey there,

A polynomial having zeros α² and β² is
(x - α²)(x - β² ) = x^2 - (α²+ β²)x + α² β².

So we need to find α²+ β² and α² β².

From x² - 1/2x -2, α+ β = 1/2 and αβ = -2.

So, α² β² = (αβ)^2 = (-2)^2 =4.
and α²+ β² = (α+ β)^2 - 2(αβ) = 1/4 +4 = 17/4.

So, a polynomial having zeros α² and β² is given by
x^2 - (17/4)x + 4.

Any multiple of this polynomial will work, too. For example, we can clear the fractions to get 4x^2 - 17x + 16.

Hope this helps!

Answer 2

Answer :-

• The value of k ↠-2

Question :-

• If α and β are the zeroes of the f(x) = x²- 6x + k, find the value of k, such that α² + β² = 40 $.$

Given :-

• α² + β² = 40

To find :-

• what is the value of k ?

Solution :-

f(x) = x²- 6x + k --------- (eq - 1 )

α and β are the zeroes of the given polynomial.

Now,

Sum of zeroes ↠ - b/a

⇒ α + β = -(-6)/1

⇒ α + β = 6

product of zeroes ↠ c/a

⇒ αβ = k

Since ,

⇒ α² + β² = 40

As we know that

(α+β)² = α² +β² +2αβ

Now , putting the values , we get

⇒ (6)²↠ 40 + 2k

⇒ 36 ↠40 + 2k

⇒ 36 - 40 ↠2k

⇒ -4 ↠2k

⇒ k ↠-4/2

⇒ k ↠-2

hence , the value of k is -2 .

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