Question

$if \: \frac{1}{ \alpha } \: and \: \alpha \: are \: the \: zeros \: of \: the \: quadric \: equation2x {}^{2} - x + 8k \: \: then \: find \: the \: value \: of \: k$
plz answer this question​

Answer 1

⇝ Given :-

$\alpha$ and $\dfrac{1}{\alpha}$ are Zeros of The Equation

2x² - x + 8k = 0.

⇝ To Find :-

• Value of k

⇝ Main Concept :-

For A Quadratic Equation of the Form :

ax² + bx + c = 0 :

$\text{Sum of Zeros} = - \dfrac{\text b}{\text a} \\ \\ \text{ Product of Zeros} = \frac{\text c}{\text a}$

⇝ Solution :-

As,α and 1/α are Zeros of The Equation 2x² - x + 8k = 0.

Hence ,

$\text{ Product of Zeros} = \dfrac{8k}{2}$

$:\longmapsto\alpha \times \dfrac{1}{ \alpha } = 4k$

$:\longmapsto1 = 4k$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf k= \frac{1}{4} } }}}$

Answer 2

Given :

α and 1/α are Zeros of The Equation 2x² - x + 8k = 0, find the value of k

To Find :

The value of k

Solution :

• p(x) = 2x² - x + 8k

Then product of zeros :

• 8k/2

• 4k

Now,

• α × 1/α = 4k

• 1 = 4k

• 4k = 1

• k = ¼

Henceforth, the value of k is 1/4 which is req. answer