Question

please help me to solve this question​

Answer 1

$\huge\sf\pink{Answer}$

☞ Value of k is -2

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$\huge\sf\blue{Given}$

✭ α & β are the zeros of x²-6x+k

✭ (α+β)²-2αβ = 40

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$\huge\sf\gray{To \:Find}$

◈ The value of k?

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$\huge\sf\purple{Steps}$

☯ $\sf\underline{\textsf{As Per the Question}}$

≫ (α+β)²-2αβ = 40 $\quad$ -eq(1)

We know that,

$\underline{\boxed{\sf\alpha+\beta =\dfrac{-b}{a}}}$

& also

$\underline{\boxed{\sf\alpha\beta = \dfrac{c}{a}}}$

On comparing x²-6x+k with ax²+bx+c

◕ a = 1

◕ b = -6

◕ c = k

Sum of zeros

➝ α+β = $\sf \dfrac{-b}{a}$

➝ α+β = $\sf \dfrac{-(-6)}{1}$

➝ α+β = $\sf \dfrac{6}{1}$

➝ α+β = 6

Product of zeros

➝ αβ = $\sf \dfrac{c}{a}$

➝ αβ = $\sf \dfrac{k}{1}$

➝ αβ = k

Substituting in eq(1)

➳ (α+β)²-2αβ = 40

➳ $\sf (6)^2-2(k) = 40$

➳ $\sf 36-2k=40$

➳ $\sf -2k=40-36$

➳ $\sf -2k=4$

➳ $\sf k=\dfrac{4}{-2}$

➳ $\sf \orange{k=-2}$

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Answer 2

jagdish of the individual and purpose is a bit more about you 6एइ