Question

If one zeroes of polynomial p(x)=(a²+10)x²+5x+7a is reciprocal of other then values of a.​

Answer 1

Answer:

a = 5,2

Step-by-step explanation:

$let \ one \ zero \ be \ k \\ and \ the \ other \ zero \ will \ be \ \frac{1}{k} \\\\ given \ polynomial,(a^{2}+10)x^{2}+5x+7a \\ it \ is \ of \ the \ form \ ax^{2}+bx+c \\ .'. \ a=(a^{2}+10) , b=5,c=7a \ \ \ \ [don't \ get \ confused \ between \ the \ two \ 'a's] \\\\ we \ know, product \ of \ roots= \frac{constant}{x^{2} \ coefficient}= \frac{c}{a}$

$k(\frac{1}{k})=\frac{7a}{a^{2}+10} \\1=\frac{7a}{a^{2}+10} \\7a=a^{2}+10 \\a^{2}-7a+10=0 \\ a^{2} -2a-5a+10=0\\a(a-2)-5(a-2)=0\\(a-2)(a-5)=0 \\ .'. \ a=5,2$

hope it helps..!