Question

if a and b are zeros of the polynomial x square-x-k such that a-b=9 find k.​

Answer 1

Answer:

k = 20

Step-by-step explanation:

We have been given the equation:-

x² - x - k = 0

Its zeroes are a and b.

Also, a - b = 9    ...(1)

The sum of zeroes of a polynomial is $\dfrac{-b}{a}$

Here, $\dfrac{-(-1)}{1} = a + b$

a + b = 1     ...(2)

Add equation (1) and (2):-

$(a - b) + (a + b) = 9 + 1$

$2a = 10$

$a = \dfrac{10}{2}$

$a = 5$

Therefore, b = -4

Product of Roots = $\dfrac{c}{a}$

$5 \times -4 = \dfrac{-k}{1}$

$-20 = -k$

$k = 20$

Therefore, the value of k is 20 !

Answer 2

p(x) = $x^{2} - x - k$

Given zeroes are a and b

And a - b = 9 (Equation 1)

Here in the polynomial,

A = 1

B = -1

C = -k

We know,

Sum of Zeroes = a + b = $\dfrac{-B}{A}$

$\implies \dfrac{-(-1)}{1} = 1$

So,

a + b = 1 (Equation 2)

Adding Equation 1and Equation 2

a - b + a + b = 9 + 1

2a = 10

a = 5

Putting value in Equation 1

a - b = 9

5 - b = 9

5 - 9 = b

b = -4

Therefore, zeroes of p(x) = $x^{2} - x - k$ are 5 and -4.

To find the value of k

Product of zeroes = a × b = 5 × (-4) = -20

We know,

Product of zeroes = $\dfrac{C}{A}$

$\implies \dfrac{-k}{1} = -20 \\ \implies k = 20$

Answer:

$\boxed{ k = 20}$