Question

If product of the zeros of the polynomial kx²+41x+42 is 7 then find the zeros of the polynomial (k-4)x²+(k+1)x+5​

Answer 1

Answer:

-1 and -5/2

Step-by-step explanation:

$Given, the \ product \ of \ the \ zeroes \ of \ the \ polynomial,kx^{2}+41x+42 \ is \ 7 \\\\ The \ relationship \ between \ product \ of \ zeroes \ and \ coefficients \ is \ given \ by, \\\\ Product \ of \ zeroes=\frac{constant}{x^{2} \ coefficient} \\ =>7=\frac{42}{k} \\\\ => k=\frac{42}{7} \\\\=>k=6 \\\\\\ find \ the \ zeroes \ of \ the \ polynomial \ (k-4)x^{2}+(k+1)x+5 \\ substitute \ the \ value \ of \ k \\\\ The \ polynomial \ becomes, \\ (6-4)x^{2}+(6+1)x+5 \\ 2x^{2}+7x+5$

$the \ quadratic \ equation => 2x^{2}+7x+5=0 \\\\ 2x^{2}+2x+5x+5=0 \\ 2x(x+1)+5(x+1)=0 \\(x+1)(2x+5)=0\\=>x+1=0 \ ; x=-1 \\=>2x+5=0 \ ; x=\frac{-5}{2} \\\\ Hence,the \ zeroes \ of \ the \ polynomial \ are \ -1 \ and \ \frac{-5}{2}$

HOPE IT HELPS..!