Question

if alpha and beta are the zeros of the polynomial 2x^2 + 5x + k . satisfying the relation Alpha square + beta square + alpha beta = 21/4 . find the value of k
should be in detailed

Answer 1

Hey!

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Zeroes are @ (alpha) and ß (beta)

Quadratic polynomial = 2x^2 + 5x + k

Where,

a = 2
b = 5
c = k

We know,

Sum of zeroes ( @ + ß ) = - b/a = -5/2

Product of zeroes ( @ß) = c/a = k / 2


Now,

Squaring both sides -:

( @ + ß ) ^2 = (-5/2)^2

@^2 + 2@ß + ß^2 = 25/4

@^2 + @ß + @ß + ß^2 = 25/4

@^2 + ß^2 + @ß + @ß = 25/4

Given,

@^2 + ß^2 + @ß = 21/4

So,

21/4 + k/2 = 25/4

k/2 = 25 / 4 - 21/4

k/2 = 25 - 21/4

k/2 = 4/4

k/2 = 1

k = 2 × 1

k = 2

Value of k = 2

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Hope it helps...!!!

Answer 2

$\textbf{Answer :}$

The given polynomial is

f (x) = 2x² + 5x + k

Since, α and β are the zeroes of f (x),

α + β = - 5/2 ...(i)

αβ = k/2 ...(ii)

Given that,

α² + β² + αβ = 21/4

⇒ (α + β)² - 2αβ + αβ = 21/4

⇒ (α + β)² - αβ = 21/4

⇒ (- 5/2)² - k/2 = 21/4, by (i) and (ii)

⇒ 25/4 - k/2 = 21/4

⇒ k/2 = 25/4 - 21/4

⇒ k/2 = (25 - 21)/4

⇒ k/2 = 4/4

⇒ k/2 = 1

⇒ k = 2

∴ The value of k is $\textbf{2}$

$\textbf{Hope it helps you.}$