Question

if one zero of the polynomial (k-1)x square+is+1 is -4, find the value of k. also find the other zero?​
please help
anyone pls help

Answer 1

Answer:

$\boxed{\text{k=5/4}} \ and \ \boxed{\text{other \ zero \ = \ -1}}$

Step-by-step explanation:

$Given \ polynomial, (k-1)x^2+kx+1 \\\\ one \ of \ the \ zeroes \ =-4 \\ let \ the \ other \ zero \ be \ y\\\\ put \ \textbf{x=-4} \\\\ (k-1)(-4)^2+k(-4)+1=0 \\16(k-1)-4k+1=0 \\ 16k-16-4k+1=0\\12k-15=0 \\ 12k=15 \\k=\frac{15}{12} \\\\ \boxed{\text{k=5/4} } \\\\ the \ quadratic \ equation, \\ =>(\frac{5}{4}-1)x^2+(\frac{5}{4})x+1 =0\\\\ => \frac{x^2}{4} + \frac{5x}{4}+1 =0 \\\\ => x^2+5x+4=0 \\\\ \\ it \ is \ of \ the \ form \ ax^2+bx+c=0 \\\\ \textbf{here; \ a=1,b=5,c=4}$

$relationship \ between \ zeroes \ and \ coefficients; \\\\ \boxed{\text{product \ of \ zeroes=c/a}} \\\\ (-4)(y) = \frac{4}{1} \\\\ -4y=4 \\\\ y=\frac{4}{-4} \\\\ y=-1 \\\\ \textbf{the \ other \ zero \ = \ -1}$

ALSO REFER :

https://brainly.in/question/1309721

$\boxed{\text{hope \ it helps}}$

Answer 2

yeh h meri bhen ...❤️❤️