If one of the zeroes of the quadratic polynomial $(k-1)x^{2}+kx+1$ is -3, then the value of k is

Question

Accepted Answer

The correct option is 1:

To solve this problem, we need to understand what a "zero" of a polynomial means. A zero of a polynomial is a value of the variable (in this case, $x$) for which the polynomial evaluates to zero.

Given the quadratic polynomial $P(x) = (k-1)x^{2}+kx+1$ and that one of its zeroes is $-3$. This means that when we substitute $x = -3$ into the polynomial, the result should be $0$.

Let's substitute $x = -3$ into the polynomial:
$P(-3) = (k-1)(-3)^{2} + k(-3) + 1 = 0$

Now, we simplify and solve for $k$:
$(k-1)(9) - 3k + 1 = 0$
$9k - 9 - 3k + 1 = 0$
Combine like terms:
$(9k - 3k) + (-9 + 1) = 0$
$6k - 8 = 0$
Add 8 to both sides:
$6k = 8$
Divide by 6:
$k = \frac{8}{6}$
Simplify the fraction:
$k = \frac{4}{3}$

Thus, the value of $k$ is $\frac{4}{3}$. This matches Option 1.

The incorrect option is 2:
This option, $\frac{-4}{3}$, would result from an arithmetic error during the simplification or solving process, such as incorrectly handling the sign of 8 or 6.

The incorrect option is 3:
This option, $\frac{2}{3}$, would result from a calculation error, possibly by incorrectly simplifying the fraction $\frac{8}{6}$ or making a mistake in solving the linear equation for $k$.

The incorrect option is 4:
This option, $\frac{-2}{3}$, would involve multiple errors in signs and/or arithmetic during the calculation, such as errors in both the signs and the simplification of the fraction.