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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$g(x)\ =\ a(x^2\ +\ 1)\ –\ x(a^2\ +\ 1)$
Given:
$g(x) = a(x^2+1) – x(a^2+1)$
To find:
Here, we have to find the zeros of g(x).
Solution:
To find the zeros of g(x), we have to put $g(x)=0$.
This implies,
$a(x^2+1) – x(a^2+1)= 0$
$ax^2+a-a^2x-x= 0$
$ax(x-a)-1(x-a)= 0$
$(ax-1)(x -a) = 0$
$ax-1=0$ and $x-a=0$
$ax = 1$ and $x = a$
$x=\frac{1}{a}$ and $x=a$
Therefore, the zeros of the quadratic equation $g(x) = a(x^2+1) – x(a^2+1)$ are $\frac{1}{a}$ and $a$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$
$= –(\frac{-(a^2+1)}{a})$
$=\frac{a^2+1}{a}$
Sum of the zeros of $g(x)=a+\frac{1}{a}=\frac{a^2+1}{a}$
Product of roots $= \frac{constant}{coefficient of x^2}$
$=\frac{a}{a}$
$= 1$
Product of the roots of $g(x)=\frac{1}{a}\times a =1$
Hence, the relationship between the zeros and their coefficients is verified.
Updated on: 10-Oct-2022
4K+ Views
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