Question

if zeroes of P ( x ) = 2 x 2 -x + K are
resiprocal of each other,
then value of k is. please give the solution with full details​

Answer 1

AnswEr:

Value of K is 2.

ExplanaTion:

Given polynomial :

• p ( x ) = 2x² - x + K

To find :

• Value of K.

SoluTion :

Let one zero be $\alpha$,

So, another be $\sf{\dfrac{1}{\alpha}}$

Comparing the given Polynomial with ax² + bx + c = 0, we get,

• a = 2
• b = -1
• c = K

Now, we know that,

$\large{\boxed{\sf{\red{Product\:of\:zeroes\:=\:\dfrac{c}{a}}}}}$

: $\implies$ $\cancel{\alpha}$ × $\dfrac{1}{\cancel{\alpha}}$ = $\sf{\dfrac{K}{2}}$

: $\implies$ 1 = $\sf{\dfrac{K}{2}}$

: $\implies$ K = 2

$\therefore$ Value of K is 2.

Answer 2

$\mathtt{ \huge{ \fbox{</strong><strong>S</strong><strong>o</strong><strong>l</strong><strong>u</strong><strong>t</strong><strong>i</strong><strong>o</strong><strong>n</strong><strong> :)}}}$

Given ,

• The polynomial is 2x² - x + k
• The zeroes of polynomial is reciprocal of each other

Let , the first zeroes of polynomial be α

Then , other = 1/α

We know that ,

The relation between product of roots and coefficient of quadratic equation is given by

$\mathtt{ \fbox{ \large{Product \: of \: roots = \frac{c}{a} }}}$

Thus ,

$\hookrightarrow \sf \alpha \times \frac{1}{ \alpha } = \frac{k}{2} \\ \\ \hookrightarrow \sf \frac{ \cancel \alpha }{ \cancel \alpha } = \frac{k}{2} \\ \\ \hookrightarrow \sf k = 2$

Hence , the value of k is 2

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