Question

If the sum of the zeroes of a polynomial p(x) = 5x

2 – 2x + (k + 1) is twice the product of its zeroes,

find k.​

Answer 1

Answer:

Let the roots be α,β

For a quadratic equation ax

2

+bx+c=0

Sum of roots α+β=k+6⋯(1)

Product of roots αβ=4k−2⋯(2)

According to question,

α+β=

2

αβ

From (1),(2)

⟹k+6=2k−1

⟹k=7

Step-by-step explanation:

hope it helps you

Answer 2

Answer:

k = 0

Step-by-step explanation:

$p(x) = 5x^2-2x + (k+1)$

Let a, b and c be the coefficients of $x^{2}$, $x$ and the constant respectively

a = 5

b = -2

c = k+1

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Let α and β be the zeroes of the polynomial

Sum of roots = α+β = -b/a = $\frac{-(-2)}{5}$

Product of roots = αβ = c/a = $\frac{k+1}{5}$

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Given that sum is twice the product

α+β = 2(αβ)

$\frac{-(-2)}{5} = 2(\frac{k+1}{5})\\\\\frac{2}{5} = \frac{2k+2}{5}\\\\2=2k + 2\\2-2 = 2k\\0 = 2k\\k=0/2\\\k=0$