Question

if 1 is one of the zero of polynomial
x^2-x+k find the value of K.
​

Answer 1

Given: $\sf x^2\ -\ x\ +\ k$; one of its zeros is 1.

To find: The value of k.

Answer:

$\sf Let's\ suppose\ ax^2\ +\ bx\ +\ c\ were\ a\ quadratic\ equation.\\\\\\The\ sum\ of\ its\ zeros\ would\ be\ given\ by\ \dfrac{-b}{a}.\\\\\\The\ product\ of\ its\ zeros\ would\ be\ given\ by\ \dfrac{c}{a}.$

From the equation we have,

• a = 1
• b = -1
• c = k

As per the question, one of the zeros is 1. Let the other be α.

Since we have the values of a and b, we can equate the sum of the zeros to 1.

$\sf 1\ +\ \alpha\ =\ \dfrac{1}{1}\\\\\\\alpha\ =\ \dfrac{1}{1}\ -\ 1\\\\\\\alpha\ =\ 0$

We now have the other zero. With both the zeros, let's now try and find the value of k.

Since c = k, let's find the product of the zeros.

$\sf 1\ \times\ 0\ =\ \dfrac{k}{1}\\\\\\0\ =\ k$

Therefore, k = 0.

Answer 2

AnswEr :

$\normalsize\bullet\:\sf\ It \: is \: given \: that \: 1 \: is \: zero \: of \\ \normalsize\sf\ polynomial \: \bf\ x^2 - x + k \: \sf\ then, \: when \: x = 1 \\ \normalsize\sf\ remainder \: is \: 0 \\ \normalsize\sf\ [i.e. P(x) = P(1)]$

$\rule{170}1$

$\scriptsize\tt{\qquad\dag\ Put \: the \: value \: of \: x = 1}$

$\normalsize\twoheadrightarrow\sf\ P(x) = x^2 - x + k$

$\normalsize\twoheadrightarrow\sf\ P(1) = (1)^2 - (1) + k$

$\scriptsize\tt{\qquad\dag\ P(x) \: is \: factor \: of \: g(x)}$

$\normalsize\twoheadrightarrow\sf\ (1)^2 - (1) + k= 0$

$\normalsize\twoheadrightarrow\sf\cancel{1} - \cancel{1} + k = 0$

$\normalsize\twoheadrightarrow\sf\ k = 0$

$\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: k \: is}{\textbf{\: 0 }}}$

$\rule{170}1$